Char array cin query (additional)

gillianbc · #1 (**permalink**) October 29th, 2004, 09:52 AM

Thanks for all you replies to my recent "Basic char array and cin query". I was really wanting to understand what was going on rather than alternative solutions hence this new thread. I'm working my way through "Beginning C++" so I don't want to introduce things that I haven't yet covered in the chapters. This query arose from the solutions to one of the exercises. In the code below (extracted directly from solutions to exercises in the book), the function header for GetData has one param that is passed by reference and one that isn't yet it still seems to work and I can't understand why.
The array gets populated by cin and its value gets passed back to main() even though it's not passed by reference - why ? What's so special about cin ? What's so special about char arrays that you don't need to use

Code:

void GetData(int& number, char& name[])

It's only cin that seems to do this. I tested this out by adding a simple assignment statement in after the cin line (hence the commented out line: name = "John") and in a debugger, you can see that name gets set to John until the end of the function but then reverts to the value input using cin.

Code:

#include <iostream.h>
void GetData(int& number, char name[])
{
    cout << "Enter a number: \n";
    cin >> number;

    if (number != 0)
    {
        cout << "And a name: \n";
        cin >> name;
        //name = "John";
    }    
}
void PutData(int number,char name[])
{
    cout    << "Thank you.  Your number and name were " << number
            << " and " << name << "\n";
}
void main()
{
    int number;
    char name[15];
    for (;;)
    {
        GetData(number, name);
        if (number==0)
            break;
        PutData(number,name);
    }    
}

Gill BC

davekw7x · #2 (**permalink**) October 29th, 2004, 11:24 AM

If you use

Code:

strcpy(name, "John");

in the function, you will see that "John" persists.

When you have a function declared like

Code:

void GetData(int& number, char name[]);

It is exactly the same as if you had

Code:

void GetData(int& number, char *name);

And as a matter of fact when you call the function, the second argument is always treated as "char *name", even if you had declared it as "char name[]". That is and always has been a part of the C language (therefore C++).

In your function, consider the statement

[code]
name = "John";
[code]

This sets the local copy of "name" to point to the C constant array, but does not affect anything in the calling program.

The strcpy() that I suggested above actually copies "John" into the array, so that the main program prints out "John".

There is nothing special about char arrays, the same is true for other arrays. In the C language there is no "string" data type; standard library functions for manipulating strings (strcat, strlen, etc.) are based on having strings represented as null-terminated arrays of char.

Regards,

Dave

gillianbc · #3 (**permalink**) October 29th, 2004, 03:03 PM

Thanks very much Dave for such a clear explanation - exactly what I needed! I just hate to move on before I've understood exactly what's going on.

Gill BC

C@uark · #4 (**permalink**) October 29th, 2004, 03:22 PM