in this code , if i want to print the sentence entered by the user then how is it possible?

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C++

#include <iostream.h>
#include <conio.h>
#include <fstream.h>

int main()
{
    char p[80];
    cout << "Enter text :";
    cin >> p;
    
    getch();
    return 0;
}

Posted 18-Aug-11 20:38pm

Ashutosh_g

Updated 19-Aug-11 10:45am

Philippe Mori

v4

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enhzflep · Answer 1 · 2011-08-18T20:43:00

Solution 2

C++

cout<<p;

Posted 18-Aug-11 20:43pm

enhzflep

Comments

Ashutosh_g 19-Aug-11 2:47am

It will only print a word of sentence, If i enter "Hello How are you" ,it will print only "Hello"

enhzflep 19-Aug-11 2:57am

Yes, but no.

It will only _input_ one word. Problm is with use of cin, not with use of cout.

char *buffer = "Hello, how are you";
cout << buffer;

Will indeed print "Hello, how are you"

I don't use cin/cout myself. Perhaps you'll find that there is a flag that ytou set for cin that will cause it to continue taking input until a line-end is found, rather than the first instance of whitespace.

enhzflep · Answer 2 · 2011-08-18T21:04:00

Solution 3

C++

const int bufLen = 1024;
char buffer[bufLen];
cin.getline(buffer, bufLen-1);
cout << buffer << endl;

Posted 18-Aug-11 21:04pm

enhzflep

Comments

Philippe Mori 19-Aug-11 16:43pm

I would recommand you to uses new C++ headers and works with string instead.

getline (with string)

వేంకటనారాయణ(venkatmakam) · Answer 3 · 2011-08-18T20:43:00

Solution 1

Like this,

cout<<p<<endl;

Posted 18-Aug-11 20:43pm

వేంకటనారాయణ(venkatmakam)

Comments

Ashutosh_g 19-Aug-11 2:46am

It will only print a word of sentence, If i enter "Hello How are you" ,it will print only "Hello"

వేంకటనారాయణ(venkatmakam) 19-Aug-11 2:58am

if you want read a line you can try like cin.getline(p, 80);check documentation of getline.

Philippe Mori 19-Aug-11 16:48pm

If real code, a constant should also be used (or the size computed using sizeof when applicable) so that the actual buffer size is defined at a single place.

Arun Parthasarathy · Answer 4 · 2011-08-22T02:16:00

Solution 4

C++

# include <iostream>
#include <conio.h>
#include <fstream>

using namespace std;
int main()
{
    
char p[80];
    cout << "Enter text :";
    cin >> p;
   cout << "the ented text is:"<< p << endl;
    
    return 0;
}

its the same code you have used.
watever u did is getting a input in variable p.
If u want tat in the console, just print the same variable thats it!

Thanks!
Arun P.

Posted 22-Aug-11 2:16am

Arun Parthasarathy

Updated 22-Aug-11 2:28am

v5

Comments

Richard MacCutchan 22-Aug-11 8:29am

Unfortunately this will not work for any input that contains more than one word; try it.

Arun Parthasarathy · Answer 5 · 2011-08-22T02:39:00

Solution 5

yes you are right!

if you want ur question to be answered, try this!

C++

#include <conio.h>
#include <fstream>
using namespace std;
int main()
{
    char p[80];
    char ch = 'y';    
while (ch == 'y' || ch == 'Y')
    {
        cout << "Enter text :";
        cin >> p;
        cout << "the ented text is" <<  p ;
        cout << "do you wish to continue y/n" << endl;
        ch = getch();
    }
    return 0;
}</fstream></conio.h>

Posted 22-Aug-11 2:39am

Arun Parthasarathy

Updated 22-Aug-11 2:45am

v2

Comments

Richard MacCutchan 22-Aug-11 9:07am

This is still wrong, why not test it properly before posting the code? The user wishes to input some text (which may include spaces) and print it out as entered.

Arun Parthasarathy 23-Aug-11 0:28am

fine. let the user reply.
rather than u testing and giving ur comments here! cant you give a proper answer if u know?