Functions

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Functions
Using functions we can structure our programs in a more modular way, accessing all
the potential that structured programming can offer to us in C++.
A function is a group of statements that is executed when it is called from some
point
of
the
program.
The
following
is
its
format:

type name ( parameter1, parameter2, ...) { statements }
where:






type is the data type specifier of the data returned by the function.
name is the identifier by which it will be possible to call the function.
parameters (as many as needed): Each parameter consists of a data type
specifier followed by an identifier, like any regular variable declaration (for
example: int x) and which acts within the function as a regular local variable.
They allow to pass arguments to the function when it is called. The different
parameters are separated by commas.
statements is the function's body. It is a block of statements surrounded by
braces { }.

// function example

The result is 8

#include <iostream>
using namespace std;
int addition (int a, int b)
{
int r;
r=a+b;
return (r);
}
int main ()
{
int z;
z = addition (5,3);
cout << "The result is " << z;
return 0;
}

Write a function to determine whether the year is a leap year or not.
#include<stdio.h>
main()
{
int leap_year(year);
int year, lp;

Prepared By Sumit Kumar Gupta, PGT Computer Science

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printf("Enter the year:");
scanf ("%d", &year);
lp=leap_year(year);
if (lp)
{
printf("nThe entered year is a leap year.");
}
else
{
printf("nThe entered year is not a leap year.");
}
}
leap_year(int y)
{
int lp;
if (y%4==0)
{
lp=1;
}
else
lp=0;
return(lp);
}

Write a general-purpose function to convert any given year into its roman
equivalent.
The following table shows the roman equivalents of decimal numbers:
Decimal:........Roman
1.....................i
5....................v
10..................x
50..................l
100................c
500...............d
1000.............m
Example:
Roman equivalent of 1988 is mdcccclxxxviii
Roman equivalent of 1525 is mdxxv
This program is a big lengthy owing to the use of Case Statements. This program can
also be rewritten using Arrays, which will reduce the length considerably.


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#include<stdio.h>
main()
{
int year;
int convert (int year);
{
printf("Note:Enter a four year digit year.nn");
printf("Enter the year that you wanna convert to Roman: " );
scanf ("%d", &year);
if (year> 1999)
{
printf("Invalid Year.Please enter again.nn");
}
}
convert(year);
}
convert(int year)
{
int i;
printf("nYear converted to Roman:");

i=(year/1000); //thousands place
if(i==1)
{
printf("m");
}
i=((year/100)%10); //hundreds place
switch (i)
{
case 1:
printf("c");
break;
case 2:
printf("cc");
break;
case 3:


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printf("ccc");
break;
case 4:
printf("cd");
break;

case 5:
printf("d");
break;
case 6:
printf("dc");
break;
case 7:
printf("dcc");
break;
case 8:
printf("dccc");
break;
case 9:
printf("dcccc"); //this part you may think is wrong..9 -> cm
break; //but i have taken a hint from the example in the question.
}
i=((year/10)%10); //tens place
switch(i)
{
case 1:
printf("x");
break;
case 2:
printf("xx");
break;
case 3:
printf("xxx");
break;


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case 4:
printf("xl");
break;
case 5:
printf("l");
break;
case 6:
printf("lx");
break;
case 7:
printf("lxx");
break;
case 8:
printf("lxxx");
break;
case 9:
printf("lxxxx"); //had it not been for this example, it would have been xc
break;
}
i=year%10; //ones place
switch(i)
{
case 1:
printf("i");
break;
case 2:
printf("ii");
break;
case 3:
printf("iii");
break;
case 4:
printf("iv");
break;


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case 5:
printf("v");
break;
case 6:
printf("vi");
break;
case 7:
printf("vii");
break;
case 8:
printf("viii");
break;
case 9:
printf("ix");
break;
}

printf ("nn");
return 0;
}

#include <stdio.h>
int main(void)
{
int i = 1, j = 2;
void exchange(int, int);
printf("main : i = %d j = %dn", i, j);
exchange(i,j);
return 0;
}
void exchange(int i, int j)
{
int t;
t = i, i = j, j = t;
printf("exchange: i = %d j = %dn", i, j);
}


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#include <stdio.h>
int main(void)
{
int i = 1, j = 2;
void exchange(int *, int *);
exchange(&i,&j);
return 0;
}
void exchange(int *ip, int *jp)
{
int t;
t = *ip, *ip = *jp, *jp = t;
printf("exchange: i = %d j = %dn", *ip, *jp);
}
The following output is produced
main : i = 1 j = 2
exchange: i = 2 j = 1
main : i = 2 j = 1
Write a function power(a,b), to calculate the value of a raised to b.
#include<stdio.h>
main()
{
int power (a,b);
int a, b, result;
printf("Enter the value of a and b:");
scanf ("%d %d", &a, &b);
result=power(a,b);
printf("%d raised to %d is %d", a, b, result);
}
power (int a, int b)
{
int calculation=1, calc;
for (calc=1; calc <=b; calc++)
{
calculation=calculation*a;
continue;
}


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return(calculation);
}
A positive integer is entered through the keyboard.
Write a function to obtain the prime factors of this number.
For example, prime factors of 24 are 2, 2, 2 and 3, whereas prime factors of
35 are 5 and 7.
#include<stdio.h>
main()
{
int number;
int prime(int number);
int primefactor(int number);

printf("Enter the number whose prime factors are to be calculated:");
scanf ("%d", &number);
primefactor(number);
}
//The following function detects a Prime number.
prime(int num)
{
int i, ifprime;

for (i=2; i<=num-1; i++)

{
if (num%i==0)
{
ifprime=0;
}
else
ifprime=1;
}
return (ifprime);
}


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//The following function prints the prime factors of a number.
primefactor(int num)
{
int factor,ifprime;
for (factor=2; factor<=num;)
{
prime(factor); //so that the factors are only prime and nothing else.
if (ifprime)
{
if (num%factor==0) //diving by all the prime numbers less than the number itself.
{
printf("%d ", factor);
num=num/factor;
continue;
}
else
{
factor++;//this cannot be made a part of the for loop
}
}
}
return 0;
}
Write a function that calculates both Area and Perimeter/ Circumference of
the Circle, whose Radius is
entered through the keyboard.
#include<stdio.h>
main()
{
int radius;
float area, perimeter;
printf("nEnter radius of a circle:");
scanf ("%d", &radius);
areaperi (radius, &area, &perimeter);
printf("Area=%f", area);
printf("nPerimeter=%f", perimeter);
}
areaperi(int r, float *a, float *p)
{


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*a=3.14*r*r;
*p=2*3.14*r;
}
Write a function which receives a float and an int from main(), finds the
product
of these two and returns the product which is printed through main().
#include<stdio.h>
main()
{
int i;
float j, prod;
float product (int x, float y);

printf("Enter the i(int) and j(float):");
scanf ("%d %f", &i, &j);
prod = product(i,j);
printf("Product:%f", prod);
}
product (int x, float y)
{
float product;
product = x*y;
return (product);
}
Write a function that receives 5 integers and returns the sum, average and
standard
deviation of these numbers. Call this function from main() and print the
results in main().
#include<stdio.h>
#include<math.h>
int calc (float a, float b, float c, float d, float e, float *sum, float *avg,float
*sd);
int main()
{
float a, b, c, d, e, sum=0.0, avg=0.0;
float sd=0.0;
printf("Enter Five Numbers:");


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scanf("%f %f %f %f %f",&a,&b,&c,&d,&e);
calc (a, b, c, d, e, &sum, &avg, &sd);

printf("nSum=%f", sum);
printf("nAverage=%f", avg);
printf("nStandard Deviation=%fn", sd);

getchar();
return 0;
}
calc (float a, float b, float c, float d, float e, float *sum, float *avg, float *sd)
{
float Calc=0.0;
*sum = a+b+c+d+e;
*avg = *sum / 5.0;
Calc += ( a - *avg) * ( a - *avg);
Calc += ( b - *avg) * ( b - *avg);
Calc += ( c - *avg) * ( c - *avg);
Calc += ( d - *avg) * ( d - *avg);
Calc += ( e - *avg) * ( e - *avg);

*sd = sqrt((double)Calc/5.0);
}
Write a recursive function to obtain the first 25 numbers of a Fibonacci
sequence. In a Fibonacci sequence the sum of two successive terms gives
the
third term. Following are the first few terms of the Fibonacci sequence:
1 1 2 3 5 8 13 21 34 55 89 ...
#include<stdio.h>
main()
{
static int prev_number=0, number=1; // static: so value is not lost
int fibonacci (int prev_number, int number);
printf ("Following are the first 25 Numbers of the Fibonacci Series:n");


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printf ("1 "); //to avoid complexity
fibonacci (prev_number,number);
}

fibonacci (int prev_number, int number)
{
static int i=1; //i is not 0, cuz 1 is already counted in main.
int fibo;

if (i==25)
{
printf ("ndone"); //stop after 25 numbers
}
else
{
fibo=prev_number+number;
prev_number=number; //important steps
number=fibo;
printf ("n%d", fibo);
i++; // increment counter
fibonacci (prev_number,number); //recursion
}
}
Write a program to calculate the factorial of a number. Use the concept of
recursion
instead of using loops.
#include<stdio.h>
main()
{
int a, fact;
printf("nEnter any number: ");
scanf ("%d", &a);
fact=rec (a);


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printf("nFactorial Value = %d", fact);
}
rec (int x)
{
int f;
if (x==1)
return (1);
else
f=x*rec(x-1);
return (f);
}
//1
main()
{
printf("nOnly stupids use c?");
display();
}
display()
{
printf("nFools too use C!");
main();
}
Answer-//1) infinite loop of both the statements

//2
main()
{
printf("nC to it that C survives.");
main();
}
Answer- //2)The message " C to it that C survives" is iterated infinitesimally!

//3
main()
{
int i=45, c;


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c=check(i);
printf("n%d", c);
}
check (int ch)
{
if (ch>=45)
return (100);
else
return (10*10);
}
Answer-//3) 100

main()
{
int i=45, c;
c=check(i*1000);
printf("n%d", c);
}
check (int ch)
{
if (ch>=40000)
return (ch/10);
else
return (10);
}
Answer--//4) 4500
We can see how the main function begins by declaring the variable z of type int.
Right after that, we see a call to a function called addition. Paying attention we will
be able to see the similarity between the structure of the call to the function and the
declaration of the function itself some code lines above:

The parameters and arguments have a clear correspondence. Within the main
function we called to addition passing two values: 5 and 3, that correspond to the int
a and int b parameters declared for function addition.
The following line of code:
return (r);


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finalizes function addition, and returns the control back to the function that called it
in the first place (in this case, main). At this moment the program follows it regular
course from the same point at which it was interrupted by the call to addition. But
additionally, because the return statement in function addition specified a value: the
content of variable r (return (r);), which at that moment had a value of 8. This value
becomes the value of evaluating the function call.

So being the value returned by a function the value given to the function call itself
when it is evaluated, the variable z will be set to the value returned by addition (5,
3), that is 8. To explain it another way, you can imagine that the call to a function
(addition (5,3)) is literally replaced by the value it returns (8).
The following line of code in main is:
cout << "The result is " << z;
That, as you may already expect, produces the printing of the result on the screen.
Scope of variables
The scope of variables declared within a function or any other inner block is
only their own function or their own block and cannot be used outside of
them. For example, in the previous example it would have been impossible
to use the variables a, b or r directly in function main since they were
variables local to function addition. Also, it would have been impossible to
use the variable z directly within function addition, since this was a variable
local to the function main.

Therefore, the scope of local variables is limited to the same block level in
which they are declared. Nevertheless, we also have the possibility to declare


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global variables; These are visible from any point of the code, inside and
outside all functions. In order to declare global variables you simply have to
declare the variable outside any function or block; that means, directly in the
body of the program.
And here is another example about functions:
// function example
#include <iostream>
int subtraction (int a, int b)
{
int r;
r=a-b;
return (r);
}

The
The
The
The

first result is 5
second result is 5
third result is 2
fourth result is 6

int main ()
{
int x=5, y=3, z;
z = subtraction (7,2);
cout << "The first result is " << z <<
'n';
cout << "The second result is " <<
subtraction (7,2) << 'n';
cout << "The third result is " <<
subtraction (x,y) << 'n';
z= 4 + subtraction (x,y);
cout << "The fourth result is " << z <<
'n';
return 0;
}
For example, the first case (that you should already know because it is the same
pattern that we have used in previous examples):
z = subtraction (7,2);
cout << "The first result is " << z;
If we replace the function call by the value it returns (i.e., 5), we would have:
z = 5;
cout << "The first result is " << z;
As well as
cout << "The second result is " << subtraction (7,2);


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has the same result as the previous call, but in this case we made the call to
subtraction directly as an insertion parameter for cout. Simply consider that the
result is the same as if we had written:
cout << "The second result is " << 5;
since 5 is the value returned by subtraction (7,2).
In the case of:
cout << "The third result is " << subtraction (x,y);
The only new thing that we introduced is that the parameters of subtraction are
variables instead of constants. That is perfectly valid. In this case the values passed
to function subtraction are the values of x and y, that are 5 and 3 respectively,
giving 2 as result.
The fourth case is more of the same. Simply note that instead of:
z = 4 + subtraction (x,y);
we could have written:
z = subtraction (x,y) + 4;
with exactly the same result. I have switched places so you can see that the
semicolon sign (;) goes at the end of the whole statement. It does not necessarily
have to go right after the function call. The explanation might be once again that you
imagine that a function can be replaced by its returned value:
z = 4 + 2;
z = 2 + 4;
Functions with no type. The use of void.
If you remember the syntax of a function declaration:
type name ( argument1, argument2 ...) statement
you will see that the declaration begins with a type, that is the type of the function
itself (i.e., the type of the datum that will be returned by the function with the return
statement). But what if we want to return no value?
Imagine that we want to make a function just to show a message on the screen. We
do not need it to return any value. In this case we should use the void type specifier
for the function. This is a special specifier that indicates absence of type.
// void function example
#include <iostream>

I'm a function!


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void printmessage ()
{
cout << "I'm a function!";
}
int main ()
{
printmessage ();
return 0;
}
void can also be used in the function's parameter list to explicitly specify that we
want the function to take no actual parameters when it is called. For example,
function printmessage could have been declared as:
void printmessage (void)
{
cout << "I'm a function!";
}
Although it is optional to specify void in the parameter list. In C++, a parameter list
can simply be left blank if we want a function with no parameters.
What you must always remember is that the format for calling a function includes
specifying its name and enclosing its parameters between parentheses. The nonexistence of parameters does not exempt us from the obligation to write the
parentheses. For that reason the call to printmessage is:
printmessage ();
The parentheses clearly indicate that this is a call to a function and not the name of a
variable or some other C++ statement. The following call would have been incorrect:
printmessage;
Arguments passed by value and by reference.
Until now, in all the functions we have seen, the arguments passed to the functions
have been passed by value. This means that when calling a function with
parameters, what we have passed to the function were copies of their values but
never the variables themselves. For example, suppose that we called our first
function addition using the following code:
int x=5, y=3, z;
z = addition ( x , y );
What we did in this case was to call to function addition passing the values of x and
y, i.e. 5 and 3 respectively, but not the variables x and y themselves.


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This way, when the function addition is called, the value of its local variables a and b
become 5 and 3 respectively, but any modification to either a or b within the function
addition will not have any effect in the values of x and y outside it, because variables
x and y were not themselves passed to the function, but only copies of their values
at the moment the function was called.
But there might be some cases where you need to manipulate from inside a function
the value of an external variable. For that purpose we can use arguments passed by
reference, as in the function duplicate of the following example:
// passing parameters by reference
#include <iostream>

x=2, y=6, z=14

void duplicate (int& a, int& b, int& c)
{
a*=2;
b*=2;
c*=2;
}
int main ()
{
int x=1, y=3, z=7;
duplicate (x, y, z);
cout << "x=" << x << ", y=" << y << ",
z=" << z;
return 0;
}
The first thing that should call your attention is that in the declaration of duplicate
the type of each parameter was followed by an ampersand sign (&). This ampersand
is what specifies that their corresponding arguments are to be passed by reference
instead of by value.
When a variable is passed by reference we are not passing a copy of its value, but
we are somehow passing the variable itself to the function and any modification that
we do to the local variables will have an effect in their counterpart variables passed
as arguments in the call to the function.

To explain it in another way, we associate a, b and c with the arguments passed on
the function call (x, y and z) and any change that we do on a within the function will
affect the value of x outside it. Any change that we do on b will affect y, and the


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same with c and z.
That is why our program's output, that shows the values stored in x, y and z after
the call to duplicate, shows the values of all the three variables of main doubled.
If when declaring the following function:
we had declared it this way:
void duplicate (int a, int b, int c)
i.e., without the ampersand signs (&), we would have not passed the variables by
reference, but a copy of their values instead, and therefore, the output on screen of
our program would have been the values of x, y and z without having been modified.
Passing by reference is also an effective way to allow a function to return more than
one value. For example, here is a function that returns the previous and next
numbers of the first parameter passed.
// more than one returning value
#include <iostream>

Previous=99, Next=101

void prevnext (int x, int& prev, int& next)
{
prev = x-1;
next = x+1;
}
int main ()
{
int x=100, y, z;
prevnext (x, y, z);
cout << "Previous=" << y << ", Next="
<< z;
return 0;
}

Default values in parameters.
When declaring a function we can specify a default value for each parameter. This
value will be used if the corresponding argument is left blank when calling to the
function. To do that, we simply have to use the assignment operator and a value for
the arguments in the function declaration. If a value for that parameter is not passed
when the function is called, the default value is used, but if a value is specified this
default value is ignored and the passed value is used instead. For example:
// default values in functions

6


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#include <iostream>

5

int divide (int a, int b=2)
{
int r;
r=a/b;
return (r);
}
int main ()
{
cout << divide (12);
cout << endl;
cout << divide (20,4);
return 0;
}
As we can see in the body of the program there are two calls to function divide. In
the first one:
divide (12)
we have only specified one argument, but the function divide allows up to two. So
the function divide has assumed that the second parameter is 2 since that is what we
have specified to happen if this parameter was not passed (notice the function
declaration, which finishes with int b=2, not just int b). Therefore the result of this
function call is 6 (12/2).
In the second call:
divide (20,4)
there are two parameters, so the default value for b (int b=2) is ignored and b takes
the value passed as argument, that is 4, making the result returned equal to 5
(20/4).
Overloaded functions.
In C++ two different functions can have the same name if their parameter types or
number are different. That means that you can give the same name to more than
one function if they have either a different number of parameters or different types
in their parameters. For example:
// overloaded function
#include <iostream>

10
2.5

int operate (int a, int b)
{
return (a*b);
}


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float operate (float a, float b)
{
return (a/b);
}
int main ()
{
int x=5,y=2;
float n=5.0,m=2.0;
cout << operate (x,y);
cout << "n";
cout << operate (n,m);
cout << "n";
return 0;
}
In this case we have defined two functions with the same name, operate, but one of
them accepts two parameters of type int and the other one accepts them of type
float. The compiler knows which one to call in each case by examining the types
passed as arguments when the function is called. If it is called with two ints as its
arguments it calls to the function that has two int parameters in its prototype and if
it is called with two floats it will call to the one which has two float parameters in its
prototype.
In the first call to operate the two arguments passed are of type int, therefore, the
function with the first prototype is called; This function returns the result of
multiplying both parameters. While the second call passes two arguments of type
float, so the function with the second prototype is called. This one has a different
behavior: it divides one parameter by the other. So the behavior of a call to operate
depends on the type of the arguments passed because the function has been
overloaded.
Notice that a function cannot be overloaded only by its return type. At least one of
its parameters must have a different type.
Arguments passed by value and by reference.
Until now, in all the functions we have seen, the arguments passed to the functions
have been passed by value. This means that when calling a function with
parameters, what we have passed to the function were copies of their values but
never the variables themselves. For example, suppose that we called our first
function addition using the following code:
int x=5, y=3, z;
z = addition ( x , y );
What we did in this case was to call to function addition passing the values of x and
y, i.e. 5 and 3 respectively, but not the variables x and y themselves.


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This way, when the function addition is called, the value of its local variables a and b
become 5 and 3 respectively, but any modification to either a or b within the function
addition will not have any effect in the values of x and y outside it, because variables
x and y were not themselves passed to the function, but only copies of their values
at the moment the function was called.
But there might be some cases where you need to manipulate from inside a function
the value of an external variable. For that purpose we can use arguments passed by
reference, as in the function duplicate of the following example:
// passing parameters by reference
#include <iostream>

x=2, y=6, z=14

{
a*=2;
b*=2;
c*=2;
}
int main ()
{
int x=1, y=3, z=7;
duplicate (x, y, z);
cout << "x=" << x << ", y=" << y << ",
z=" << z;
return 0;
}
The first thing that should call your attention is that in the declaration of duplicate
the type of each parameter was followed by an ampersand sign (&). This ampersand
is what specifies that their corresponding arguments are to be passed by

reference instead of by value.
When a variable is passed by reference we are not passing a copy of its value, but
we are somehow passing the variable itself to the function and any modification that
we do to the local variables will have an effect in their counterpart variables passed
as arguments in the call to the function.

To explain it in another way, we associate a, b and c with the arguments passed on
the function call (x, y and z) and any change that we do on a within the function will
affect the value of x outside it. Any change that we do on b will affect y, and the
same with c and z.


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That is why our program's output, that shows the values stored in x, y and z after
the call to duplicate, shows the values of all the three variables of main doubled.
If when declaring the following function:
we had declared it this way:
void duplicate (int a, int b, int c)
i.e., without the ampersand signs (&), we would have not passed the variables by
reference, but a copy of their values instead, and therefore, the output on screen of
our program would have been the values of x, y and z without having been modified.
Passing by reference is also an effective way to allow a function to return more than
one value. For example, here is a function that returns the previous and next
numbers of the first parameter passed.
// more than one returning value
#include <iostream>

Previous=99, Next=101

void prevnext (int x, int& prev, int& next)
{
prev = x-1;
next = x+1;
}
int main ()
{
int x=100, y, z;
prevnext (x, y, z);
cout << "Previous=" << y << ", Next="
<< z;
return 0;
}
Default values in parameters.
When declaring a function we can specify a default value for each parameter. This
value will be used if the corresponding argument is left blank when calling to the
function. To do that, we simply have to use the assignment operator and a value for
the arguments in the function declaration. If a value for that parameter is not passed
when the function is called, the default value is used, but if a value is specified this
default value is ignored and the passed value is used instead. For example:
// default values in functions
#include <iostream>

6
5

int divide (int a, int b=2)


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{
int r;
r=a/b;
return (r);
}
int main ()
{
cout << divide (12);
cout << endl;
cout << divide (20,4);
return 0;
}
As we can see in the body of the program there are two calls to function divide. In
the first one:
divide (12)
we have only specified one argument, but the function divide allows up to two. So
the function divide has assumed that the second parameter is 2 since that is what we
have specified to happen if this parameter was not passed (notice the function
declaration, which finishes with int b=2, not just int b). Therefore the result of this
function call is 6 (12/2).
In the second call:
divide (20,4)
there are two parameters, so the default value for b (int b=2) is ignored and b takes
the value passed as argument, that is 4, making the result returned equal to 5
(20/4).
Overloaded functions.
In C++ two different functions can have the same name if their parameter types or
number are different. That means that you can give the same name to more than
one function if they have either a different number of parameters or different types
in their parameters. For example:
// overloaded function
#include <iostream>

10
2.5

int operate (int a, int b)
{
return (a*b);
}
float operate (float a, float b)
{
return (a/b);
}


of 33
int main ()
{
int x=5,y=2;
float n=5.0,m=2.0;
cout << operate (x,y);
cout << "n";
cout << operate (n,m);
cout << "n";
return 0;
}
Recursive Functions:Recursive is the property that functions have to be called themselves
For Example;- n!=n(n-1)*(n-2)*(n-3)……..*1

Functions QuestionsFind the output of the following program
class Sample
{
public:
int *ptr;
Sample(int i)
{
ptr = new int(i);
}
~Sample()
{
delete ptr;
}
void PrintVal()
{
cout << "The value is " << *ptr;
}
};
void SomeFunc(Sample x)
{
cout << "Say i am in someFunc " << endl;
}
int main()
{
Sample s1= 10;
SomeFunc(s1);
s1.PrintVal();
}
Answer:
Say i am in someFunc
What do you mean by inline function?


of 33
The idea behind inline functions is to insert the code of a called function at the point
where the function is called. If done carefully, this can improve the applications
Performance in exchange for increased compile time and possibly (but not always)
an increase in the size of the generated binary executables.
What do you mean by binding of data and functions?
Encapsulation.
What is abstraction?
Abstraction is of the process of hiding unwanted details from the user.
What is encapsulation?
Packaging an object’s variables within its methods is called encapsulation.
What is friend function?
As the name suggests, the function acts as a friend to a class. As a friend of a class,
it can access its private and protected members. A friend function is not a member of
the class. But it must be listed in the class definition.
What are virtual functions?
A virtual function allows derived classes to replace the implementation provided by
the base class. The compiler makes sure the replacement is always called whenever
the object in question is actually of the derived class, even if the object is accessed
by a base pointer rather than a derived pointer. This allows algorithms in the base
class to be replaced in the derived class, even if users don't know about the derived
class.
What does extern mean in a function declaration?
Using extern in a function declaration we can make a function such that it can used
outside the file in which it is defined.
An extern variable, function definition, or declaration also makes the described
variable or function usable by the succeeding part of the current source file. This
declaration does not replace the definition. The declaration is used to describe the
variable that is externally defined.
If a declaration for an identifier already exists at file scope, any extern declaration of
the same identifier found within a block refers to that same object. If no other
declaration for the identifier exists at file scope, the identifier has external linkage.
How do I declare an array of N pointers to functions returning pointers to
functions returning pointers to characters?
Answer1
If you want the code to be even slightly readable, you will use typedefs.
typedef char* (*functiontype_one)(void);
typedef functiontype_one (*functiontype_two)(void);
functiontype_two myarray[N]; //assuming N is a const integral

Answer2


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char* (* (*a[N])())()
Here a is that array. And according to question no function will not take any
parameter value.
What does extern mean in a function declaration?
It tells the compiler that a variable or a function exists, even if the compiler hasn’t
yet seen it in the file currently being compiled. This variable or function may be
defined in another file or further down in the current file.
How do I initialize a pointer to a function?
This is the way to initialize a pointer to a function
void fun(int a)
{
}
void main()
{
void (*fp)(int);
fp=fun;
fp(1);
}
Write a fucntion that will reverse a string.
char *strrev(char *s)
{
int i = 0, len = strlen(s);
char *str;
if ((str = (char *)malloc(len+1)) == NULL)
/*cannot allocate memory */
err_num = 2;
return (str);
}
while(len)
str[i++]=s[–len];
str[i] = NULL;
return (str);
}


of 33

Object Oriented Programming
Data Abstraction:-is the act of representing the essential features without including
the background details and explanations.
Data Hiding:-is a related concept to data abstraction. Unessential features or
background details are hidden from the world.
Data Encapsulation:-is the wrapping up of data and associated functions in one
single until,while implementing abstraction at the same time.
Objects:-represents an identifiable entity with some characteristics and behaviour.
What do you mean by inheritance?
Inheritance is the process of creating new classes, called derived classes, from
existing classes or base classes. The derived class inherits all the capabilities of the
base class, but can add embellishments and refinements of its own.
Base Class:-is the class, which gets inherited from other classes. It is also called
Super class.
Derived class is the class which inherits from other classes. It is also called
Subclass.
What is polymorphism? Explain with an example?
"Poly" means "many" and "morph" means "form". Polymorphism is the ability of an
object (or reference) to assume (be replaced by) or become many different forms of
Object.
Example: function overloading, function overriding, virtual functions. Another
example can be a plus
‘+’ sign, used for adding two integers or for using it to concatenate two strings.
OPP is approach for writing software in which data and behaviour are packaged
together as classes whose instances are objects.
A class is a named software representation for an abstraction.
What is function overloading and operator overloading?
Function overloading: C++ enables several functions of the same name to be
defined, as long as these functions have different sets of parameters (at least as far
as their types are concerned). This capability is called function overloading. When an
overloaded function is called, the C++ compiler selects the proper function by
examining the number, types and order of the arguments in the call.
Function overloading is commonly used to create several functions of the same name
that perform similar tasks but on different data types.
Operator overloading allows existing C++ operators to be redefined so that they
work on objects of user-defined classes. Overloaded operators are syntactic sugar for
Equivalent function calls. They form a pleasant facade that doesn't add anything
fundamental to the language (but they can improve understandability and reduce
Maintenance costs).

Very Short Answer Questions:-1-2 Marks
Fill in the blanks;
(i)………….. is a named collection of attributes and behaviour related to modeling a
given entity for some particular purpose?


of 33
(ii) …………..is a programming approach where data and behaviour are encapsulated
to gether as…………
(iii) An instance of a class is called ……………
(iv) A class enforces data hiding through …………and……………members.
(v) Polymorphism is implemented in c++ through ……..and……………
Answer:- 1) OPP, Classes 2) Abstraction 3) Object 4) Private, Protected 5) Virtual
functions, overloading.
What is abstract class?
Ans:- Abstract class is a class that defines an interface, but does not necessary
provide implementation for all its member functions. No objects of an abstract class
exit i.e, it cannot be instantited.
What is concrete class?
Ans:- A Concrete class is a derived class of an abstract class that implements all the
missing functionality . A concrete class can be instantiated i.e. its objects can be
created.
What is the importance of the function overloading?
Ans:- Function overloading not only implements polymorphism but also reduces
number of comparisons in a program and thereby makes the program run faster.
How objects implemented in c++?
Ans:- The object is implemented in software terms as follows:(1) Characteristics / attributes are implemented through member variables or
data items of the object.
(2) Behaviour is implemented through member functions called method.
(3) It is given a unique name to give it identify.
Why do you think function overloading must be a part of an object oriented
language?
Ans: - A Function overloading must be a part of an object oriented language as it is
the feature that implements polymorphism in an object oriented language. That is
the ability of an object to behave differently circumstances can effectively be
implemented in programming through function overloading.
Find the syntax error, if any in the following program.
Include<iostream.h>
Void main()
Int R; w=90;
While w>60
{
R=W-50;
Switch(W)
{
20: cout<<”Lower range”<<endl;
30: cout<<”Middle range”<<endl;
40: cout<<”Higher range”<<endl;
}
}
}
#include<iostream.h>
void main()


of 33
{
int X,Y,
cin>>X;
for(Y=0;Y<10;Y++)
{ (X==Y)
cout<<Y+X;
else
cout>>4;
}
}
Include<iostream.h>
Void main()
{
int P[ ]= {90,60,30,10}; Q;Number=4;
Q=9;
For[int i=Number-1;>=0;i--]
Switch(i)
{
case0:
case 3: cout>>P[i]*Q<<endl; break;
case 1:
case 2: cout<<p[i]+q;
}
}
Give the output of the following program:#include<iostream.h>
int global=10;
void sunil(int &x, int y)
{
x=x-y;y=x*10;
cout<<x<<”,”<<y<<’n’;
}
void main()
{
int global=7;
sunil(::global,global);
cout<<global<<”,”<<::global<<’n’;
sunil(global,::global);
cout<<global<<”,”<<::global<<’n;
}
Ans:- 3,30
7,3
4,40
4,3

Give the output of the following program:void main()
{
char*name=”ProOcessoR”
for(int x=0,x<strlen(name);x++
{
if(islower(name[x]))


of 33
name(x)=toupper(name[x]);
else
if(isupper(name[x]))
if(x%2==0)tolower
name[x]=(name[x]);
else
name(x)=name[x-1];
}
puts(name);
}
Ans:- pRo OEesOr

What is function overloading? Use the concept of the function overloading
to compute area of rectangle (given length of two sides), area of a triangle
(given length of the three sides using Hero’s formula) and area of circle
(given length of a radius).
#include<iostream.h>
#include<conio.h>
#include<math.h>
float area(float a, float b)
{
return a*b;
float area(float a, float b, float c)
{
float s, ar;
s=(a+b+c)/2;
ar=sqrt(s*(s-a)*(s-b)*(s-c));
return ar;
}
float area(float a)
{
return 3.1423*a*a;}
int main()
{
clrscr();
int choice, S1,S2,S3;
float ar;
do
{
cout<<”n Area Main Menun”;
cout<<”n Rectanglen”;
cout<<”n Trianglen”;
cout<<”n Circlen”;
cout<<”n Exitn”;
cout<<”n Enter your choice(1-4):-”;
cin>>choice; cout<<endl;
switch(choice)


of 33
{
case 1:

case 2:

case 2:

case 4:

cout<<”Enter Length & Breadthn”;
cin>>S1>>S2; ar=area(S1,S2);
cout<<”The Area is<<ar<<endl”;
break;
cout<<”Enter 3 sidesn”;
cin>>S1>>S2>>S3; ar=area(S1,S2,S3);
break;
cout<<”Enter Radiusn”;
cin>>S1; ar=area(S1);

break;
break;

default: cout<<”Wrong choice!n”;
};
} while(choice>)&&choice<4);
return 0;
}


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