One foot equals 12 inches. Write an assembly language function int to.pdf
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One foot equals 12 inches. Write an assembly language function int toInches(int feet) that accepts a length written in feet as an argument and returns this length written in inches. Can you help me write an ASSEMBLY LANGUAGE? I am using Seggar, here is my C code:/********************************* main() Function description Application entry point. int toInches(int feet) {float i; i=feet*(12); return i;} Can you explain the assembly language? I am a beginner. Thank you. Solution the assambly languge code to convert feet to inches is mention below, toInches(int): push rbp mov rbp, rsp mov DWORD PTR [rbp-20], edi mov edx, DWORD PTR [rbp-20] mov eax, edx add eax, eax add eax, edx sal eax, 2 pxor xmm0, xmm0 cvtsi2ss xmm0, eax movss DWORD PTR [rbp-4], xmm0 movss xmm0, DWORD PTR [rbp-4] cvttss2si eax, xmm0 pop rbp ret.
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![One foot equals 12 inches. Write an assembly language function int toInches(int feet) that
accepts a length written in feet as an argument and returns this length written in inches. Can you
help me write an ASSEMBLY LANGUAGE? I am using Seggar, here is my C
code:/********************************* main() Function description Application entry
point. int toInches(int feet) {float i; i=feet*(12); return i;} Can you explain the assembly
language? I am a beginner. Thank you.
Solution
the assambly languge code to convert feet to inches is mention below,
toInches(int):
push rbp
mov rbp, rsp
mov DWORD PTR [rbp-20], edi
mov edx, DWORD PTR [rbp-20]
mov eax, edx
add eax, eax
add eax, edx
sal eax, 2
pxor xmm0, xmm0
cvtsi2ss xmm0, eax
movss DWORD PTR [rbp-4], xmm0
movss xmm0, DWORD PTR [rbp-4]
cvttss2si eax, xmm0
pop rbp
ret](https://image.slidesharecdn.com/onefootequals12inches-230708122139-14d49f80/85/One-foot-equals-12-inches-Write-an-assembly-language-function-int-to-pdf-1-320.jpg)
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please help implement the following in C. below are instructions and partial code below that,
step one is completed in the image.
double getMean(const int *arr, int size) {
//TODO: implement
}
int getMin(const int *arr, int size) {
//TODO: implement
}
int getIndexOfMin(const int *arr, int size) {
//TODO: implement
}
int getMax(const int *arr, int size) {
//TODO: implement
}
int getIndexOfMax(const int *arr, int size) {
//TODO: implement
}
int * filterThreshold(const int *arr, int size, int threshold, int *resultSize) {
//TODO: implement
}
int **createMultiplicationTable(int n, int m) {
//TODO: implement
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be able to utilize others. - double getMean(const int *arr, int size) - computes the average of
elements in arr - int getmin (const int *arr, int size) - returns the minimum element in arr - int
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Translate the following CC++ code into MIPS Assembly Codevoid ch.pdf
Translate the following C/C++ code into MIPS Assembly Code:
void chico(int *X, int Y, int Z)
{
*X = Y/4 - Z*10 + *X*8;
}
int main()
{
int J, K, L, M;
cin >> J, K, L;
chico ( &J, K, L);
M = J - (K + L);
cout << M;
return 0;
}
Solution
Program:
#include
using namespace std;
void chico(int *X, int Y, int Z)
{
*X = Y/4 - Z*10 + *X*8;
}
int main()
{
int J, K, L, M;
cin >> J,K,L;
chico ( &J, K, L);
M = J - (K + L);
cout<<\"value\"<< M;
return 0;
}
The above code returns the following Mips code :
chico(int*, int, int):
push rbp
mov rbp, rsp
mov QWORD PTR [rbp-8], rdi
mov DWORD PTR [rbp-12], esi
mov DWORD PTR [rbp-16], edx
mov eax, DWORD PTR [rbp-12]
lea edx, [rax+3]
test eax, eax
cmovs eax, edx
sar eax, 2
mov ecx, eax
mov edx, DWORD PTR [rbp-16]
mov eax, edx
sal eax, 2
add eax, edx
add eax, eax
neg eax
lea edx, [rcx+rax]
mov rax, QWORD PTR [rbp-8]
mov eax, DWORD PTR [rax]
sal eax, 3
add edx, eax
mov rax, QWORD PTR [rbp-8]
mov DWORD PTR [rax], edx
nop
pop rbp
ret
.LC0:
.string \"value\"
main:
push rbp
mov rbp, rsp
sub rsp, 16
lea rax, [rbp-16]
mov rsi, rax
mov edi, OFFSET FLAT:std::cin
call std::basic_istream >::operator>>(int&)
mov edx, DWORD PTR [rbp-8]
mov ecx, DWORD PTR [rbp-4]
lea rax, [rbp-16]
mov esi, ecx
mov rdi, rax
call chico(int*, int, int)
mov eax, DWORD PTR [rbp-16]
mov ecx, DWORD PTR [rbp-4]
mov edx, DWORD PTR [rbp-8]
add edx, ecx
sub eax, edx
mov DWORD PTR [rbp-12], eax
mov esi, OFFSET FLAT:.LC0
mov edi, OFFSET FLAT:std::cout
call std::basic_ostream >& std::operator<< >(std::basic_ostream >&, char const*)
mov rdx, rax
mov eax, DWORD PTR [rbp-12]
mov esi, eax
mov rdi, rdx
call std::basic_ostream >::operator<<(int)
mov eax, 0
leave
ret
__static_initialization_and_destruction_0(int, int):
push rbp
mov rbp, rsp
sub rsp, 16
mov DWORD PTR [rbp-4], edi
mov DWORD PTR [rbp-8], esi
cmp DWORD PTR [rbp-4], 1
jne .L6
cmp DWORD PTR [rbp-8], 65535
jne .L6
mov edi, OFFSET FLAT:std::__ioinit
call std::ios_base::Init::Init()
mov edx, OFFSET FLAT:__dso_handle
mov esi, OFFSET FLAT:std::__ioinit
mov edi, OFFSET FLAT:std::ios_base::Init::~Init()
call __cxa_atexit
.L6:
nop
leave
ret
push rbp
mov rbp, rsp
mov esi, 65535
mov edi, 1
call __static_initialization_and_destruction_0(int, int)
pop rbp
ret
The above includes all the code that is written even for braces represented above..
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double getMean(const int *arr, int size) {
//TODO: implement
}
int getMin(const int *arr, int size) {
//TODO: implement
}
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//TODO: implement
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//TODO: implement
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//TODO: implement
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//TODO: implement
}
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//TODO: implement
}
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void chico(int *X, int Y, int Z)
{
*X = Y/4 - Z*10 + *X*8;
}
int main()
{
int J, K, L, M;
cin >> J, K, L;
chico ( &J, K, L);
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return 0;
}
Solution
Program:
#include
using namespace std;
void chico(int *X, int Y, int Z)
{
*X = Y/4 - Z*10 + *X*8;
}
int main()
{
int J, K, L, M;
cin >> J,K,L;
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The above code returns the following Mips code :
chico(int*, int, int):
push rbp
mov rbp, rsp
mov QWORD PTR [rbp-8], rdi
mov DWORD PTR [rbp-12], esi
mov DWORD PTR [rbp-16], edx
mov eax, DWORD PTR [rbp-12]
lea edx, [rax+3]
test eax, eax
cmovs eax, edx
sar eax, 2
mov ecx, eax
mov edx, DWORD PTR [rbp-16]
mov eax, edx
sal eax, 2
add eax, edx
add eax, eax
neg eax
lea edx, [rcx+rax]
mov rax, QWORD PTR [rbp-8]
mov eax, DWORD PTR [rax]
sal eax, 3
add edx, eax
mov rax, QWORD PTR [rbp-8]
mov DWORD PTR [rax], edx
nop
pop rbp
ret
.LC0:
.string \"value\"
main:
push rbp
mov rbp, rsp
sub rsp, 16
lea rax, [rbp-16]
mov rsi, rax
mov edi, OFFSET FLAT:std::cin
call std::basic_istream >::operator>>(int&)
mov edx, DWORD PTR [rbp-8]
mov ecx, DWORD PTR [rbp-4]
lea rax, [rbp-16]
mov esi, ecx
mov rdi, rax
call chico(int*, int, int)
mov eax, DWORD PTR [rbp-16]
mov ecx, DWORD PTR [rbp-4]
mov edx, DWORD PTR [rbp-8]
add edx, ecx
sub eax, edx
mov DWORD PTR [rbp-12], eax
mov esi, OFFSET FLAT:.LC0
mov edi, OFFSET FLAT:std::cout
call std::basic_ostream >& std::operator<< >(std::basic_ostream >&, char const*)
mov rdx, rax
mov eax, DWORD PTR [rbp-12]
mov esi, eax
mov rdi, rdx
call std::basic_ostream >::operator<<(int)
mov eax, 0
leave
ret
__static_initialization_and_destruction_0(int, int):
push rbp
mov rbp, rsp
sub rsp, 16
mov DWORD PTR [rbp-4], edi
mov DWORD PTR [rbp-8], esi
cmp DWORD PTR [rbp-4], 1
jne .L6
cmp DWORD PTR [rbp-8], 65535
jne .L6
mov edi, OFFSET FLAT:std::__ioinit
call std::ios_base::Init::Init()
mov edx, OFFSET FLAT:__dso_handle
mov esi, OFFSET FLAT:std::__ioinit
mov edi, OFFSET FLAT:std::ios_base::Init::~Init()
call __cxa_atexit
.L6:
nop
leave
ret
push rbp
mov rbp, rsp
mov esi, 65535
mov edi, 1
call __static_initialization_and_destruction_0(int, int)
pop rbp
ret
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#include
using namespace std;
void *routineName(void *arg)
{
// TODO: Add code that implements
// the thread\'s functionality
cout << \"Thread is running...\" << endl;
return 0;
}
int main()
{
pthread_t id;
int rc;
int ints;
for(int x; x<10; x++)
{
cout << \"Enter Integer: \" <>
}
rc = pthread_create(&id, NULL, routineName, NULL);
if (rc){
cout << \"ERROR; return code from pthread_create() is \" << rc << endl;
return -1;
}
pthread_exit(0);
}
Solution
#include
#include
#include
using namespace std;
int ints[10];
//Method to count negatives in the input array
void *countNegatives(void*){
int count = 0;
for(int i=0;i<10;i++){
if(ints[i]<0){
count++;
}
}
cout<<\"Number of negatives: \"<=0;i--){
cout<> ints[x];
}
//Creating pthread to countNegatives
rc = pthread_create(&t1, NULL, countNegatives, NULL);
if(rc){
cout << \"Error:unable to create thread,\" << rc << endl;
}
//Adding delay
sleep(1);
//Creating pthread to compute average
rc = pthread_create(&t2, NULL, average, NULL);
if(rc){
cout << \"Error:unable to create thread,\" << rc << endl;
}
//Adding delay
sleep(1);
//Creating pthread to print in reverse
rc = pthread_create(&t3, NULL, reverse, NULL);
if(rc){
cout << \"Error:unable to create thread,\" << rc << endl;
}
//Adding delay
sleep(1);
void* status;
//waiting for t1 to join
rc = pthread_join(t1,&status);
if(rc){
cout << \"Error:unable to join,\" << rc << endl;
}
//waiting for t2 to join
rc = pthread_join(t2,&status);
if(rc){
cout << \"Error:unable to join,\" << rc << endl;
}
//waiting for t3 to join
rc = pthread_join(t3,&status);
if(rc){
cout << \"Error:unable to join,\" << rc << endl;
}
pthread_exit(0);
}.
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drainRate (rate at which water drains from pool in gallons per minute)
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and currentGallons = 25000 to pool1\'s attributes.
Calculate the totalGallons for pool1 and pool2.
Use the output methods to print the attribute\'s values for both pools.
Use the member method to calculate the time it will take to completely fill pool1. Use get
methods to output the current gallons and additional gallons needed to fill the pool, along with
the fill time.
Use the member method to calculate the time it will take to completly drain pool2. Use the get
method to output the number of gallons along with the drain time.
Modify the currentGallons in pool1 if water is added for 5 hours
Modify the currentGallons in pool2 if water is drained for 10 hours
Use the output methods to print the attribute\'s values for both pools
output:
Pool 1 - Swimming Pool Attributes
Length: 0.0 feet
Width: 0.0 feet
Depth: 0.0 feet
Fill rate: 0.0 gallons/minute
Drain rate: 0.0 gallons/minute
Current gallons: 0.0 gallons
Total capacity: 0.0 gallons
Pool 2 - Swimming Pool Attributes:
Length: 16.0 feet
Width: 32.0 feet
Depth: 4.5 feet
Fill rate:4.0 gallons/minute
Drain rate: 8.0 gallons/minute
Current gallons: 12000.0 gallons
Total capacity: 0.0 gallons
Pool 1 - Swimming Pool Attributes:
Length: 20.0 feet
Width: 40.0 feet
Depth: 5.5 feet
Fill rate: 4.5 gallons/minute
Drain rate: 10.0 gallons/minute
Current gallons: 25000.0 gallons
Total capacity: 33000.0 gallons
Pool 2 - Swimming Pool Attributes:
Length: 16.0f.
I need the code for a Java programming project. Allow the user to ty.pdf
I need the code for a Java programming project. Allow the user to type in a integer or floating
point decimal number. calculate and display the binary equivalent. Use 2’s complement for
negative numbers. For floating point numbers use simplified format
Solution
Answer:
import java.io.*;
class ComplementProgram
{
public static void main(String args[]) throws IOException
{
int read[]=new int[32];
int input1,input2,i=0,carryflag=0,j=0,k=0,m,n;
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
input1=Integer.parseInt(br.readLine());
input2=input1;
i=31;
while(input1!=0)
{
read[i]=Math.abs(input1%2);
input1=input1/2;
i=i-1;
}
while(i>=0)
{
read[i]=0;
i=i-1;
}
System.out.println(\"\");
for(k=0;k<32;k++)
System.out.print(read[k]);
if(input2<0)
{
for(i=0;i<32;i++)
{
if(read[i]==1)
read[i]=0;
else
read[i]=1;
}
carryflag=1;
for(i=31;i>=0;i--)
{
if(read[i]==1 && carryflag==1)
{
carryflag=1;
read[i]=0;
}
else if(carryflag==1)
{
read[i]=1;
carryflag=0;
break;
}
}
}
System.out.println(\"\");
if(input2<0)
{
System.out.println(\"The 2\'s Complement of a positive number is:\");
for(k=0;k<32;k++)
System.out.print(read[k]);
}
else
System.out.println(\"For this number there is no 2\'s complement\");
}
}.
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Modify this code to use multiple threads with the same data
1.Modify the main function to implement a loop that reads 10 integers from the console (user
input) and stores these numbers in a one-dimensional (1D) array (this code will go right after the
comment that says “Add code to perform any needed initialization or to process user input”).
You should use a global array for this.
2.Implement a separatepthread function function for each one of the following operations:
a.Count and print out how many of the entered numbers are negative. This function must be
named countNegatives
b.Calculate and print the average value of all the numbers entered. This function must be named
average
c.Print the numbers in reverse order from the order in which they were entered. This function
must be named reverse
3.Modify the main function to create one pthread for each one of the functions that you
implemented in (3) above (this code will go between the comment that says “TODO: Modify
according to assignment requirements” and the “if (rc)” check).
Compile your program and run it several times. If the output of your program is garbled, you
may need to add a small delay in between creating the next thread.
#include
#include
using namespace std;
void *routineName(void *arg)
{
// TODO: Add code that implements
// the thread\'s functionality
cout << \"Thread is running...\" << endl;
return 0;
}
int main()
{
pthread_t id;
int rc;
int ints;
for(int x; x<10; x++)
{
cout << \"Enter Integer: \" <>
}
rc = pthread_create(&id, NULL, routineName, NULL);
if (rc){
cout << \"ERROR; return code from pthread_create() is \" << rc << endl;
return -1;
}
pthread_exit(0);
}
Solution
#include
#include
#include
using namespace std;
int ints[10];
//Method to count negatives in the input array
void *countNegatives(void*){
int count = 0;
for(int i=0;i<10;i++){
if(ints[i]<0){
count++;
}
}
cout<<\"Number of negatives: \"<=0;i--){
cout<> ints[x];
}
//Creating pthread to countNegatives
rc = pthread_create(&t1, NULL, countNegatives, NULL);
if(rc){
cout << \"Error:unable to create thread,\" << rc << endl;
}
//Adding delay
sleep(1);
//Creating pthread to compute average
rc = pthread_create(&t2, NULL, average, NULL);
if(rc){
cout << \"Error:unable to create thread,\" << rc << endl;
}
//Adding delay
sleep(1);
//Creating pthread to print in reverse
rc = pthread_create(&t3, NULL, reverse, NULL);
if(rc){
cout << \"Error:unable to create thread,\" << rc << endl;
}
//Adding delay
sleep(1);
void* status;
//waiting for t1 to join
rc = pthread_join(t1,&status);
if(rc){
cout << \"Error:unable to join,\" << rc << endl;
}
//waiting for t2 to join
rc = pthread_join(t2,&status);
if(rc){
cout << \"Error:unable to join,\" << rc << endl;
}
//waiting for t3 to join
rc = pthread_join(t3,&status);
if(rc){
cout << \"Error:unable to join,\" << rc << endl;
}
pthread_exit(0);
}.
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java programming 5th edition chapter 8 user defined classes and ADTs.pdf
java programming 5th edition chapter 8 user defined classes and ADTs use with strings.
1. Write the definition of a class called SwimmingPool, to implement the properties of a
swimming pool.
instance variables:
length (in feet)
width (in feet)
depth (in feet)
fillRate (rate at which water fills pool in gallons per minute)
drainRate (rate at which water drains from pool in gallons per minute)
currentGallons (current number of gallons to fill the pool)
totalGallons (number of gallons to fill the pool)
Constructor methods:
default constructor
values constructor
Member methods:
separate \'set\' methods for all instance variables, except totalGallons, which will always be
calculated
separate \'get\' methods for all instance variables
Determine number of gallons the pool will hold (hint: the total gallons a rectangular pool will
hold is the pool\'s volume * 7.5)
The time needed to completely fill the pool (based on the current gallons)
The time needed to completely drain the pool (based on the current gallons)
Modify the gallons if water is added or drained for a specific number of minutes
Output the values of all instance variables
2. Create a client that will do each of the following:
Instantiate a pool1 object with the default constructor.
Instantiate a pool2 object with the values constructor, use length = 16, width = 32, depth = 4.5,
fillRate = 4, drainRate = 8, and currentGallons = 12000.
Use the output method to print all attribute values for both pools.
Use the set methods to assign length = 20, width = 40, depth = 5.5, fillrate = 4.5, drainrate = 10,
and currentGallons = 25000 to pool1\'s attributes.
Calculate the totalGallons for pool1 and pool2.
Use the output methods to print the attribute\'s values for both pools.
Use the member method to calculate the time it will take to completely fill pool1. Use get
methods to output the current gallons and additional gallons needed to fill the pool, along with
the fill time.
Use the member method to calculate the time it will take to completly drain pool2. Use the get
method to output the number of gallons along with the drain time.
Modify the currentGallons in pool1 if water is added for 5 hours
Modify the currentGallons in pool2 if water is drained for 10 hours
Use the output methods to print the attribute\'s values for both pools
output:
Pool 1 - Swimming Pool Attributes
Length: 0.0 feet
Width: 0.0 feet
Depth: 0.0 feet
Fill rate: 0.0 gallons/minute
Drain rate: 0.0 gallons/minute
Current gallons: 0.0 gallons
Total capacity: 0.0 gallons
Pool 2 - Swimming Pool Attributes:
Length: 16.0 feet
Width: 32.0 feet
Depth: 4.5 feet
Fill rate:4.0 gallons/minute
Drain rate: 8.0 gallons/minute
Current gallons: 12000.0 gallons
Total capacity: 0.0 gallons
Pool 1 - Swimming Pool Attributes:
Length: 20.0 feet
Width: 40.0 feet
Depth: 5.5 feet
Fill rate: 4.5 gallons/minute
Drain rate: 10.0 gallons/minute
Current gallons: 25000.0 gallons
Total capacity: 33000.0 gallons
Pool 2 - Swimming Pool Attributes:
Length: 16.0f.
I need the code for a Java programming project. Allow the user to ty.pdf
I need the code for a Java programming project. Allow the user to type in a integer or floating
point decimal number. calculate and display the binary equivalent. Use 2’s complement for
negative numbers. For floating point numbers use simplified format
Solution
Answer:
import java.io.*;
class ComplementProgram
{
public static void main(String args[]) throws IOException
{
int read[]=new int[32];
int input1,input2,i=0,carryflag=0,j=0,k=0,m,n;
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
input1=Integer.parseInt(br.readLine());
input2=input1;
i=31;
while(input1!=0)
{
read[i]=Math.abs(input1%2);
input1=input1/2;
i=i-1;
}
while(i>=0)
{
read[i]=0;
i=i-1;
}
System.out.println(\"\");
for(k=0;k<32;k++)
System.out.print(read[k]);
if(input2<0)
{
for(i=0;i<32;i++)
{
if(read[i]==1)
read[i]=0;
else
read[i]=1;
}
carryflag=1;
for(i=31;i>=0;i--)
{
if(read[i]==1 && carryflag==1)
{
carryflag=1;
read[i]=0;
}
else if(carryflag==1)
{
read[i]=1;
carryflag=0;
break;
}
}
}
System.out.println(\"\");
if(input2<0)
{
System.out.println(\"The 2\'s Complement of a positive number is:\");
for(k=0;k<32;k++)
System.out.print(read[k]);
}
else
System.out.println(\"For this number there is no 2\'s complement\");
}
}.
How each of the main themes of The Fifth Discipline are related to e.pdf
How each of the main themes of The Fifth Discipline are related to each other ?
Solution
The five discipline are as follows-system thinking,personal mastery,mental model,building
shared vision ,team learning.
Each discipline consists of principle ,proposition or concepts.so its hardly matters from where to
start.each discipline has its impact on other discipline .each made up of set of tools and practices
for building the sustainning leadership nature in organisation..
Hi,BST question I was wondering why my insert function works p.pdf
Hi,
BST question
I was wondering why my insert function works perfectly to insert file data in the tree but doesnt
when i insert single one
here is my function
int BSTInsert(BST *tree, BSTKey key, BSTValue value)
{
BSTNode *prev = NULL;
BSTNode *cur = tree->Root;
//
// first we search the tree to see if it already contains key:
//
while (cur != NULL)
{
if (BSTCompareKeys(key, cur->Key) == 0) // already in tree, failed:
return 0;
else if (BSTCompareKeys(key, cur->Key) < 0) // smaller, go left:
{
prev = cur;
cur = cur->Left;
}
else // larger, go right:
{
prev = cur;
cur = cur->Right;
}
}
//
// If we get here, tree does not contain key, so insert new node
// where we fell out of tree:
//
BSTNode *T = (BSTNode *)malloc(sizeof(BSTNode));
T->Key = key;
T->Value = value;
T->Left = NULL;
T->Right = NULL;
//
// link T where we fell out of tree -- after prev:
//
if (prev == NULL) // tree is empty, insert @ root:
{
tree->Root = T;
}
else if (BSTCompareKeys(key, prev->Key) < 0) // smaller, insert to left:
{
prev->Left = T;
}
else // larger, insert to right:
{
prev->Right = T;
}
tree->Count++;
return 1; // success:
}
here i insert data from file
fgets(buf, sizeof(buf), fp);
buf[strcspn(buf, \"\ \ \")] = \'\\0\';
token = strtok(buf, \"\\t\");
y=atoi(token);
value.Y = y;
token=strtok(NULL, \"\ \");
strcpy(x,token);
value.X = (char *)malloc((strlen(x) + 1) * sizeof(char));
strcpy(value.X, x);
BSTInsert(tree, value.X, value);
while (fgets(buf, sizeof(buf), fp) != NULL)
{
buf[strcspn(buf, \"\ \ \")] = \'\\0\';
token = strtok(buf, \"\\t\");
y=atoi(token);
value.Y = y;
token=strtok(NULL, \"\ \");
strcpy(x,token);
value.X = (char *)malloc((strlen(x) + 1) * sizeof(char));
strcpy(value.X, x);
BSTInsert(tree, value.X, value);
}
here i insert data from keyboard
BSTValue value1;
long long int weight;
char part2[512];
int part2size = sizeof(part2) / sizeof(part2[0]);
//
// add weight text
//
scanf(\"%lld %s\", &weight, phrase);
fgets(part2, part2size, stdin);
part2[strcspn(part2, \"\ \ \")] = \'\\0\'; // strip EOL char(s):
strcat(phrase, part2);
value1.Y = weight;
value1.X = (char *)malloc((strlen(phrase) + 1) * sizeof(char));
strcpy(value1.X, phrase);
BSTInsert(tree, value1.X, value1);
if(BSTInsert(tree, value1.X, value1)==1){
printf(\"**Added\ \");
}
else
{
printf(\"**Not dded\ \");
}
thanks in advance
Solution
int BSTInsert(BST *tree, BSTKey key, BSTValue value)
{
BSTNode *prev = NULL;
BSTNode *cur = tree->Root;
//
// first we search the tree to see if it already contains key:
//
while (cur != NULL)
{
if (BSTCompareKeys(key, cur->Key) == 0) // already in tree, failed:
return 0;
else if (BSTCompareKeys(key, cur->Key) < 0) // smaller, go left:
{
prev = cur;
cur = cur->Left;
}
else // larger, go right:
{
prev = cur;
cur = cur->Right;
}
}
//
// If we get here, tree does not contain key, so insert new node
// where we fell out of tree:
//
BSTNode *T = (BSTNode *)malloc(sizeof(BSTNode));
T->Key = key;
T->Value = value;
T->Left = NULL;
T->Right = NULL;
//
// link T where w.
Favor Composition does what and howSolutionWe know what is in.pdf
Favor Composition does what and how?
Solution
We know what is inheritance, inheritance is an oops feature, acquiring existing properties form
the existing class isknow as inheritance. when a Field’s type is a class, the field will hold a
reference to another object, thus creating an association relationship between them. Without
getting into the nuances of the difference between simple association, aggregation, and
composition, let’s intuitively define composition as when the class uses another object to provide
some or all of its functionality.
Favor composition over inheritance is a one of the popular object-oriented design principles,
which helps to create flexible and maintainable code in Java and other object-oriented languages
A super class reference variable can refer to an instance of sub class. By using composition, you
don’t get this behavior, but still it offers a lot more to tilde the balance in its side.
Inheritance and composition relationships are also referred as IS-A and HAS-A relationships.
Favor composition give.
Explain how character displacement can influence the speciation p.pdf
Explain how character displacement can influence the speciation process.
Explain how character displacement can influence the speciation process.
Solution
Disruptive selection leads to sympatric speciation. Sympatric speciation defined as the evolution
of novel species with spatial isolation or distribution of the common similar ancestral species
without any continental drift of separation that means from the species of same geographical
area. In this kind of speciation, we can observe no continental separation for interbreeding
between species and thereby no divergence in the resultant species. Disruptive selection
predominantly is depending on severe competition between same niches for resource partitioning
& character displacement that influence the speciation process. This competition result in
sexually reproductive isolation lastly sympatric speciation. Heritability based on predation
pressure and allelic frequency to represent the selection and evolutionary change. It has
illustrated that the same evolutionary features have associated with normal populations in which
a few genetic traits have passed from one generation to another generation based on reproductive
fitness and environmental adaptation. It has profoundly found from these simulations that any
heritable genetic trait for character displacement, which meticulously increases species
biological, phenotypic and reproductive fitness, referred as an “adaptation” acquired by that
species to the environment..
astronomy One Nineteenth Century observation which convinced many.pdf
astronomy One Nineteenth Century observation which convinced many of the existence of at
least plant life on Man was the behavior of the polar caps and dark areas. What was this
behavior? An attempted invasion during the 1930\'s which was broadcast on the radio. The
northern polar cap grows during the northern winter while the dork areas in the northern
hemisphere begin to radiate x-ray and radio waves. The polar caps are connected to the dark
areas by a network of canal v The dark areas in the northern hemisphere grow during the
northern summer as the northern polar cap recedes.
Solution
D) is the answeer.
The planet Mars has two permanent polar ice caps. During a pole\'s winter, it lies in continuous
darkness, chilling the surface
https://en.wikipedia.org/wiki/Martian_polar_ice_caps.
A) For bipyridine (bipy) determine the maximum number of coordinatio.pdf
A) For bipyridine (bipy) determine the maximum number of coordination sites that the ligand
can occupy on a single metal ion.
B) For bipyridine (bipy) determine the number of donor atoms in the ligand.
C) For bipyridine (bipy) determine the type of donor atoms in the ligand. The donor atoms will
be either N, C, or H.
Solution
a) two
b) 2
c) N.
Describe two (2) attributes of a globally aware EngineerEngineering.pdf
Describe two (2) attributes of a globally aware Engineer/Engineering Associate.
Solution
1. To Express and Exhibit an understanding of
Technology Fundamentals and engineering science
political, social, and economic evaluations
Product lifecycle Phases and its stages (introduction, growth, maturity, and decline)
Planning and Impacts of Project on various stakeholder groups (project team members, project
promoter, project client, customers, etc.)
2. Acquaring the skill and talent to visualize both critically and productively, to visualize both
cooperatively and independently..
C# This program lets you pick from a choice of 8 items Circle, Squ.pdf
C#: This program lets you pick from a choice of 8 items: Circle, Square, Ellipse, Pie, Filled
Circle, Filled Square, Filled Ellipse, and Filled Pie. When Square is selected, it draws and
displays a square . I want to add 1 button to change the color of the sharp which I want to create.
How can I get it. ajal ComboBox Test Square Square Ellipse Pie Filled Circle Filled Square
Filled Ellipse Filled Pie
Solution
using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Text;
using System.Windows.Forms;
namespace graphics
{
public partial class Form1 : Form
{
public Form1()
{
InitializeComponent();
}
Pen red = new Pen(Color.Black);
private void comboBox1_SelectedIndexChanged(object sender, EventArgs e)
{
panel1.Refresh();
if (comboBox1.SelectedIndex == 2)
{
SolidBrush sb = new SolidBrush(Color.White);
Graphics g = panel1.CreateGraphics();
g.DrawEllipse(red,20, 20, 220, 90);
}
if (comboBox1.SelectedIndex == 1)
{
//panel1.Refresh();
SolidBrush sb = new SolidBrush(Color.White);
Graphics g = panel1.CreateGraphics();
g.DrawRectangle(red, 20, 20, 20, 20);
}
if (comboBox1.SelectedIndex == 3)
{
SolidBrush sb = new SolidBrush(Color.White);
Graphics g = panel1.CreateGraphics();
g.DrawPie(red, 100, 70, 50, 50, 60, 20);
}
if (comboBox1.SelectedIndex == 0)
{
int radius=50;
SolidBrush sb = new SolidBrush(Color.White);
Graphics g = panel1.CreateGraphics();
g.DrawEllipse(red, 100 - radius, 100 - radius, radius + radius, radius + radius);
}
if (comboBox1.SelectedIndex == 4)
{
int radius = 50;
SolidBrush sb = new SolidBrush(Color.Red);
Graphics g = panel1.CreateGraphics();
g.DrawEllipse(red, 100 - radius, 100 - radius, radius + radius, radius + radius);
g.FillEllipse(sb, 100 - radius, 100 - radius, radius + radius, radius + radius);
}
if (comboBox1.SelectedIndex == 5)
{
SolidBrush sb = new SolidBrush(Color.Red);
Graphics g = panel1.CreateGraphics();
g.DrawRectangle(red, 100, 100, 100, 100);
g.FillRectangle(sb, 100, 100, 100, 100);
}
if (comboBox1.SelectedIndex == 6)
{
SolidBrush sb = new SolidBrush(Color.Red);
Graphics g = panel1.CreateGraphics();
g.DrawEllipse(red, 20, 20, 220, 90);
g.FillEllipse(sb, 20, 20, 220, 90);
}
if (comboBox1.SelectedIndex == 7)
{
SolidBrush sb = new SolidBrush(Color.Red);
Graphics g = panel1.CreateGraphics();
g.DrawPie(red, 100, 70, 50, 50, 60, 20);
g.FillPie(sb, 100, 70, 50, 50, 60, 20);
}
}
}
}.
“Our communication capabilities have raced ahead of our communicatio.pdf
“Our communication capabilities have raced ahead of our communication abilities. Our
communication is getting worse”
Base on the above quotation:
1. Do you find this true in the workplace?
2. Support your claim with examples.
Solution
Communication
Communication is the process of exchanging the thoughts by one person with another person or
group of persons. The communication needs to be effective so that the followers can understand
it properly and can understand the subject.
The communication needs to be in a common language between the speaker and listeners so that
both can have effective interaction between them and understand the topic effectively. Effective
communication can only make the listener to understand the topic. If the language is
inappropriate then the listeners will not be able to understand it properly. If the speaker will not
be able to speak properly, then he will not be able to communicate effectively and he will not be
able to explain the topic the listeners group. The message or information cannot be transferred
without the effective communication between the listeners and the speaker. Effective
communication can help to understand the topics in one go; otherwise it will take long time for
the learners to understand the topic in absence of proper communication.
Communication Capabilities vs Abilities: Our communication capabilities have raced ahead of
our communication abilities. Our communication is getting worse” This is the right statement as
our communication capabilities means the process of communications improved to an higher
level with the help of latest technologies in communications like email, video-calling, voice-
calling, animated learning/communicative models, branding and marketing means etc. All these
communication processes are improved a large scale. Our communication abilities are reduced to
very high level due to lack of proper communications between the people or group of people.
Thus people used to communicate less thus their communication abilities are reduced while the
way of communication i.e. capabilities improved to a large extent due to advancement in
communication technologies.
This is a true statement from the workplace point of view. At the work place the communication
between the employees and between the manager and employees are reduced. The one of the
reason for reduction in communication is due to the reduction in the abilities of communication
among employees.
Employees are not skilled in communication. This effects the reduction in communication.
Examples:.
18. The density of a noble gas at STP is 1.784 gL. Identify the nobl.pdf
18. The density of a noble gas at STP is 1.784 g/L. Identify the noble gas. a. Xe b. Ne c. Kr d.
Ar
Solution
Ar is the nobel gas which has density of 1.784 g/dm³
(1dm³=1L)
Nobel gasses have weak forces of attraction,so their molecules remains at large distance than
other elements
So in a definite volume they have small mass
So their density is very low..
You have identified a mutation in E. coli K-12 that causes it to bec.pdf
You have identified a mutation in E. coli K-12 that causes it to become sensitive to the the
antibiotic azithromycin (azi). You sequence the chromosome of the aziS strain and localize the
mutation to genX and near the cysABC operon that encodes the only genes required for cysteine
biosynthesis (see diagram). In order to study the impact of the genX* mutation in a pathogenic
strain, you would like to move this allele into a uropathogenic E. coli (UPEC) strain background.
Design a strategy to move your genX* mutation from E. coli K-12 into the UPEC strain using
the strains and plasmids described below. Use point form to describe your strategy. Include a
description of all conditions used for selections and/or screens and be clear, but brief about how
you will carry out each step. Hint: your answer will require the use of transformation,
transposition, and transduction and should include a protocol for a genetic screen for a
transposon mutant & the use of generalized transduction to move alleles between strains (9
marks): •Suicide plasmid pMUT: plasmid that encodes the transposon Tn5 & ampicillin
resistance; can only replicate at 30oC, not 37oC. the Tn5 element contains a gene that confers a
kanamycin resistant phenotype (KnR) •E. coli K-12 genX* : E. coli K-12 strain carrying your
genX* mutation that confers azithromycin sensitivity •UPEC wild-type strain •Generalized
transducing phage P1, which can infect both E. coli K-12 and UPEC
(all information is given: need to design an experiment with non-mobilizable/ suicide vectors/
transposons/ selection process etc). UPEC chromosome genX cySA cysB *cySA, B, Care only
genes required for biosynthesis of cysteine EC Ampicillin Ori pMUT Tn5 (KnR
Solution
1. Transduction and Experimental steps for transduction:-
Transduction is the heritable transfer of bacterial DNA from one cell (donor, here it is E Coli
K12) to another (recipient, EColi, UPEC) by a bacteriophage. Tansducing bacteriophage
particles are formed in donor bacterial cells during phage development. Generalized transducing
phage particles carry a random fragment of host chromosomal DNA approximately the same
length as P1 DNA. The phage particles completely lack DNA originating from the phage
genome and contain instead oly Bacterila DNA sequesnces. New genotypes in the recipient cells
result from homoogous recombination, which can lead to the replacement of a recipient gene by
an allele acquired from the donor genome via the transducing phage.
a. Ecoli K 12 (Host cell) infected by phage
b. Host cell DNA broken down into smaller pieces and proteins and phage DNA synthesized
c. Bacterial host DNA is packaged in some viral capsids that are released through lysis of
bacterial cell.
d. Transducing phage with host DNA infest new recipient cell.
e. Recombinant new cell contain mixture of donor DNA and its own DNA.
2. Transformation:- It is genetic alteration of a cell resulting from the direct uptake and
incorporation of exogenous genetic material fr.
Why are the EROI values for U.S. ethanol fuel from corn different in.pdf
Why are the EROI values for U.S. ethanol fuel from corn different in the studies performed by
Shapouri, et al. (2004), Pimentel and Patzek (2005), and Farrell, et al. (2006)? Explain in terms
of input energy and co-product energy.
Solution
Why are the EROI values for U.S. ethanol fuel from corn different in the studies performed by
Shapouri, et al. (2004), Pimentel and Patzek (2005), and Farrell, et al. (2006):
Over the past decade there has been considerable debate on corn ethanol, most focused on
whether it is a net energy yielder. The argument is generally that “if the Energy Return on
Investment (EROI) of corn ethanol is positive then it should be pursued. On one side are
Pimentel (2003) and Patzek (2004) who claim that corn ethanol has an EROI below one energy
unit returned per energy unit invested, and on the other side are a number of studies claiming that
the EROI is positive, reported variously as between 1.08 and 1.45.Even with numerous
publications on this issue, disagreement remains as to whether corn ethanol is a net energy
yielder.
We believe that focus within the literature on whether or not corn ethanol yields a positive net
energy gain has diverted attention from more fundamental issues.
Following is a brief description of some of these issues:
First, none of the major studies of the EROI of corn ethanol account for statistical error within
their analysis. Error is associated with all measurements, and we should expect there to be error
associated with EROI as well. Yet each of Farrell et al. (2006), Wang et al. (2007), Patzek
(2004), Pimentel (2003), and Shapouri et al. (2002) fail to report even general error statistics
associated with their calculation of EROI. Considering that the range of published values for the
EROI of corn ethanol is so small (from 0.8 to 1.5) one would expect that even a relatively small
amount of error could be meaningful.
Second, most analyses to date, including those referenced above, use optimal values for corn
yield, fertilizer, and irrigation, despite the fact that each of these have large geographical
variation. Because of this they fail to represent the variable nature of corn production across
space, and by extension the subsequent variability in the EROI of corn ethanol. Our spatial
analysis addressed this issue by examining the impacts of the natural geographic variability of
corn inputs and yields on the EROI of corn ethanol production within the U.S.
We performed four major analyses in this research:
The rst was a meta-error analysis, in which we quantied the error associated with the calculation
of EROI of corn ethanol based on various estimates of the energy inputs and outputs found in the
literature. This analysis was based on the five main studies in corn ethanol: Wang et al. (1997),
Shapouri et al. (2002), Pimentel (2003), Patzek (2004), and Farrell et al. (2006).
Second was a spatial analysis of the EROI of corn ethanol.
Third was a sensitivity analysis; wherein we assess the degre.
Which of the following are parts of a bond Choose one or more A. t.pdf
Which of the following are parts of a bond? Choose one or more: A. the interest rate B. the
default risk C. the date of maturity D. the par value E. the name of the lender F. the name of the
borrower
Solution
The following is a part of a bond:
C. the date of maturity.
Every bond has mainly three parts which is mentioned in the bond.
Face value, maturity date and coupon rate..
When we decide to make something compensable we assert its value.pdf
When we decide to make something \"compensable\" we assert its value, relative to other factors.
For your response, consider the following two questions: 1) As you observe our society what are
some of the key compensable factors that affect who gets paid what? 2) Given those factors,
what can you surmise/deduce are some of the key assumptions (about work and pay or about
society in general) underlying the relative distribution of values?
Solution
These 10 elements describe the specific job requirements in terms of compensable factors. The
primary compensable factors that affects are:
1. Experience- How long should the incumbent have workrd in this job or in closely related jobs
to be fully qualified? It is important that the experience be within or outside the organisation.
2. Education- What does the job require in terms of formal schooling, training, certificate or
knowledge of a specialized field?
3. Responsibility- Is the employee responsible for the safety of other employees or for the loss or
damage to tools, materials or equipement?
4. Complexity of duties- Does the job require the incumbent to show judgement and initiative or
to make independent decisions?
5. Supervision received- How closely does the incumbent\'s immediate suoervisor or manager
check his or her work?
6. Supervision exercised- How many people does the incumbent\'s supervise, directly or
indirectly?
7. Consequences of error- If the incumbent made an error, what dollar loss would be likely to
result?
8. Working conditions- Is there anything in the work environment that is unusually hazardous or
uncomfortable?
9. Mental, Physical and visual demands- What degree of concentration is required? Are there
special physical demands?
10. Confidential data- To what extent is the incumbent responsible for confidential information?
To what extent are integrity and discretion important?.
What is the role of ports for global commerce and why is that roe im.pdf
What is the role of ports for global commerce and why is that roe important? Is our current port
infrastructure sufficient?
Solution
Ports play major role in the global commerce of the country as major imports and exports are
done from the port. It is the source of transport for many countries . Heavy material cannot be
taken in and out of the country without these ports. The ports are also major part of the economy
of the country and employement source. Many people work here and extract their living. The
ports of the country are well developed and have every latest technology incorporated and are
considered one of the best . However they need to be updated and maintained inorder to continue
the effciency..
What are the stages of team development Why is it important for a s.pdf
What are the stages of team development? Why is it important for a supervisor to understand
these stages?
Solution
Team can be defined as group of people who work together with a high degree of
interdependence in order to complete a task or project. The team will exhibit highest level of
performance if they have the same purpose, there is an open communication within the team,
members trust each other, create positive working environment, resolve conflict effectively,
flexible and adaptable to change and focus on learning new things.
The stages of team formation are as follows: -
Forming –
Storming –
Norming –
Performing –
Adjourning –
Effective team development is important for supervisor because: -.
Two numbers together add to 300. One number is twice the size of the.pdf
Two numbers together add to 300. One number is twice the size of the other. What are the two
numbers?
A. 25; 50
B. 50; 100
C. 75; 150
D. 100; 200
Solution
15. D. Set this up as an equation. A number plus 2 times that number is 300.
x + 2x = 300
Solve for x.
3x=300
X=100
So one number is 100 and the other is twice that, or 200..
This Question 2 Pts The primary goal of financial accounting is to p.pdf
This Question: 2 Pts The primary goal of financial accounting is to provide information to O A.
potential investors. OB. government regulators. OC. creditors. O D. all of the above. Click to
select your answer
Solution
D. all of the above
The primary goal of financial accounting is to provide information for governmental regulators,
creditors and potential investors..
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