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Managing console


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This is a precise ppt on managing consoles I/O operations. i hope you will find it good and useful.

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Managing console

  1. 1. Submitted by:- Namita Pandey2011BTechece020 Shiva Johari
  2. 2.  Introduction Streams & Stream Classes Unformatted Input Output Operations Formatted Console Input Output Operation Formatting Flags, Bit fields and setf() Designing Our Own Manipulators
  3. 3.  Managing Console I/O Operations INPUT & OUTPUT C++ supports a rich set of I/O operations C++ uses the concept of stream & stream classes
  5. 5.  C++ contains a hierarchy of classes that are used to definevarious streams ios INPUT OUTPUT POINTER istream streambuf ostream iostreamIstream_withassign Iostream_withassign Ostream_withassign
  6. 6.  Overloaded operators >> and << get() and put() functions getline() and write() functions
  7. 7. C++ supports a number of features which can be used for formatting the output. These features include :- ios class functions Manipulators User-defined Manipulators
  8. 8. FUNCTION TASK Width()  To specify the required field size for displaying the output value Precision()  To specify the digits to be displayed after decimal point of a float value Fill()  To specify a character that is used to fill the unused portion of a field Setf()  To specify format flags that can control the form of output display Unsetf()  To clear the flags specified
  9. 9. MANIPULATORS EQUIVALENT IOS FUNCTION setw()  width() setprecision()  precision() setfill()  fill() setiosflags()  setf() resetiosflags()  unset()
  10. 10. cout.width(5); 5 4 3 1 2cout<<543<<12<<“n”;cout.width(5);`cout<<543;cout.width(5); 5 4 3 1 2cout<<12<<“n”;
  11. 11. #include<iostream.h> cout.width(8); cout << cost[i];int main(){ int value = items[i] * int item[4] = cost[i]; {10,8,12,15}; cout.width(15); int cost[4] = cout << value <<“n”; {75,100,60,99}; cout.width(5); sum =sum + value; cout << “ITEMS”; } cout.width(8); cout << “n Grand cout << “COST”; Total = “; cout.width(15); cout << value<<“n”; cout.width(2); sum =sum + value; cout << sum <<“n”; int sum = 0; for(int i=0; i<4;i++) return 0; { cout.width(5); } cout << items[i];
  12. 12. ITEMS COST TOTAL VALUE 10 15 150 8 100 800 12 60 720 15 99 1485
  13. 13. cout.precision(3); 1.141cout<<sqrt(2)<<“n”; 3.142cout<<3.14159<<“n”;cout<<2.50032<<“n”; 2.5
  14. 14. #include<iostream.h>#include<conio.h>void main(){ float pi=22.0/7.0; int I; cout<<“Value of pi :n “; for(i=1;i<=10;i++) { cout.width(i+1); cout.precision(i); cout<<pi<<“n”; }}
  15. 15. Value of pi :
  16. 16. cout.fill(‘*’);cout.width(10);cout<<5250<<“n”; * * * * * * 5 2 5 0
  17. 17. #include<iostream.h> cout<<“n Paddling Changednn”;#include<conio.h> cout.fill(‘#’);void main() cout.width(15);{ cout<<12.345678<<“n”;cout.fill(‘<‘); return 0;cout.precision(3); }for(int n=1;n<=6;n++){ cout.width(5); cout<<n; cout.width(10); cout<<1.0/float(n)<<“n”; if(n==3) cout.fill(‘>’);}
  18. 18. <<<<1<<<<<<<<<1<<<<2<<<<<<<0.5<<<<3<<<<<<<0.3>>>>4>>>>>>0.25>>>>5>>>>>>>0.2>>>>6>>>>>0.167Paddling Changed#########12.346
  19. 19. arg1 - formatting flags defined in the class iosarg2 - it specifies the group to which the formatting flags belongs
  20. 20. FORMAT REQUIRED FLAG (ARG1) BIT-FIELD (ARG2) Left-justified output ios::left ios::adjustfield Right-justified output ios::right ios::adjustfieldPadding after sign or base ios::internal ios::adjustfield Indicator (like +##20) Scientific Notation ios::scientific ios::floatfield Fixed Point notation ios::fixed ios::floatfield Decimal Base ios::dec ios::basefield
  21. 21. #include<conio.h>#include<iostream.h>main(){cout.setf(ios::fixed, ios::floatfield);float x=1234.67 ;cout<<x<<endl;cout.setf(ios::scientific, ios::floatfield);x=.123467 ;cout<<x;getch();}
  22. 22. 1234.6682341.234672e-01
  23. 23. #include<iostream.h>#include<conio.h>void main(){ int num; cout<<“enter an integer value”; cin>>num; cout<<“The hexadecimal, octal and decimal representation is : ”; cout.setf(ios::hex, ios::basefield) cout<<num<<“, “; cout.setf(ios::oct, ios::basefield) cout<<num<<“, “; cout.setf(ios::dec, ios::basefield) cout<<“ and “<<num<<“ respectively”;}
  24. 24. Enter an integer value : 92The hexadecimal, octal and decimalrepresentation of 92 is: 5c, 134 and 92respectively.
  25. 25. ostream & manipulator (ostream & output){………………………… (code)……………return output;}
  26. 26.  We have taken all the basic ideas about theconcepts from the book “OBJECT ORIENTEDTECHNIQUES” – by E. Balagurusamy Images are made in Ms- Paint & every thing is accompanied by ideas of ourown
  27. 27. The function of istream class is to :a) inherit the properties of iosb) Declares input functions such as get(), getline(), read() etc.c) Contains overloaded extraction operator >>d) All of the above
  28. 28. The function of streambuf is to :a) provides an interface to physical devices through buffersb) Can’t act as a base for filebuf class used for ios filesc) Declares constants and functions that are necessary for handling formatted i/p and o/p operationsd) None of the above
  29. 29. A stream is a sequence of ___________.a) Bytesb) Filesc) Manipulatorsd) None of the above
  30. 30. Which are the member functions of ios class :a) precision()b) width()c) fill()d) All the above
  31. 31. What will be the output of following : cout.fill(‘*’); cout.precision(3); cout.setf(ios::internal, ios::adjustfield); cout.setf(ios::scientific, ios::floatfield); cout.width(15); cout<<-12.34567<<“n”;-******1.235e+01 (.A B.) -*****1.235e+01-*****.1235e+02 -*********1.236 (.C D.)
  32. 32. a) The __________ operator is overloaded in the istream classa) Insertionb) >>c) <<d) None of the above
  33. 33. Which Class is needed to be virtual in this case : a.) iostream b.) ios c.) istream or ostream d.) no one is required ios INPUT OUTPUT POINTER istream streambuf ostream iostreamIstream_withassign Iostream_withassign Ostream_withassign
  34. 34. Q8.The header file iomanip can be used in place ofiostream ??Q9. programmer can’t define a manipulator that couldrepresent a set of formatted functions ??
  35. 35. What is the default precision value ?? a.) 0 b.) 4c.) 6 d.) 3