Thread: Using cin to read a name into 1 string

Doesn't this work?

Code:

string firstName, lastName;
cin >>firstName >>lastName;

^ yes, but he wants it in one string... to clarify speedy 5's answer, do this:

Code:

char name[256];
cin.getline(name,256,'\n');

using getline takes the '\n' out of the istream, and using cin.get() leaves the '\n' in the ifstream. I find getline() more handy because I usually don't want the character in there... the middle number can be left out (cin.getline(name,'\n'))... that just makes sure that the system stops reading in after reading that amount of chars without reaching the term char...

I didn't use the string datatype in my example because I don't know much about it, and I don't think that's the way you do it with strings...

Originally posted by major_small
^ yes, but he wants it in one string... to clarify speedy 5's answer, do this:

Code:

char name[256];
cin.getline(name,256,'\n');

using getline takes the '\n' out of the istream, and using cin.get() leaves the '\n' in the ifstream. I find getline() more handy because I usually don't want the character in there... the middle number can be left out (cin.getline(name,'\n'))... that just makes sure that the system stops reading in after reading that amount of chars without reaching the term char...

Not being an expert in getline() I have this question. If you leave out the len parameter, isn't the function analogous to gets() in C? IOW:

Code:

char name[10]
cin.getline(name,'\n');

what will happen if you type in 20 or 30 characters? Is C++ smart enough to stop at 10? If not, leaving out the middle number is extremely dangerous.

C++ will stop at 10 characters in this instance. You can see this by printing name[10] to the screen after input. I would always use the second paramater however just to be safe.

Originally posted by /Muad'Dib\
C++ will stop at 10 characters in this instance. You can see this by printing name[10] to the screen after input. I would always use the second paramater however just to be safe.

name[10] doesn't exist, what good is printing it? What would you see?

Test program:

Code:

#include <iostream>
using namespace std;
int main ()
{
    char mn[10];
    cin.getline(mn, '\n');
    cout << mn << endl;
    cout << "10=[" << mn[10] << "]" << endl;
    return 0;
}

Execution:

Code:

d:\>xp
abcdefghijkl
abcdefghi
10=[ ]

d:\>

Scary. No indication that mn[10] is non-existant. But it did stop reading after 9 characters.

I don't think it's exactly dangerous to your machine... it will just try to save it in a place that doesn't exist... it's still an error though... if you try to read past the end of an array either your compiler will tell you it doesn't exist, or your system will have to decide what to do with it... sometimes it'll just put a random character in the stream...

how about using fgets


#define SIZE 100
char array[SIZE];


fgets(array,SIZE,stdin);


and it works perfect for me..

>Is C++ smart enough to stop at 10? If not, leaving out the middle number is extremely dangerous.
In your example, where '\n' has the numeric value of 10, yes. Leaving out the middle argument is dangerous, and it doesn't do what most people would expect. It takes the decimal value of the delimiter charater and uses that as the size argument. So the call:

Code:

char name[15];
cin.getline ( name, '\n' );

Will read up to int('\n') - 1 characters, or stop when '\n' is reached as that happens to be the default for this overload of getline.

Remarkable coincidence.

>Remarkable coincidence.

You can say that again ... I thought that getline() call looked mighty suspicious.

if you want to wait for the user to hit enter to accept the full string you could use the code below with the header conio.h

Code:

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

string input(){
     string text = "";
     char get = getch(); 
     while( get != '\r'){     // \r is the enter key
          cout<<get;
          text += get;
          get = getch();
          }
     return text;
}

Hope this helped