Returning Variables

I know I asked this question about a year ago but I came upon a simple solution that is probably common knowledge.

Code:

---

    int retvar()
    {
          char retvar[]="retvar";
          return retvar;
    }

    int main()
    {
      retvar();
      cout << retvar;
      }

---

whats wrong? when i put that in "int/void main it doesn't return the variable. And in advance, how would i go about returning multiple variables?

don't you have to define the function in order to return a value?

The function:

Code:

---

int retvar()
    {
          char retvar[]="retvar";
          return retvar;
    }

---

declares an int return value then tries to return a char*. It also tries to return a char* which will point to nothingness - retvar is destroyed at the end of the function.

In addition to this, main has no way of knowing about retvar. It has not been declared or referenced or anything anywhere.

The following should accomplish what you are trying to do - I hope :)

Code:

---

#include <iostream.h>

void retvar( char[] charArray, int arraySize )
{
  charArray = "cat";
}

int main( )
{
  char array[ 100 ];  //100 characters in an array
  retvar( array, 100 );
  cout << array << endl;
}

---

Arrays are passed by reference so there is no need to return the array.

This probably isn't very clear, just shout if you have questions :)

Quote:

---

Originally posted by endo

The following should accomplish what you are trying to do - I hope :)

---

Nope.

I'll make a try:

Code:

---

char* retvar( char* charArray, int arraySize )
{
  strncpy(charArray,"cat",arraySize-1);
  charArray[arraySize-1]='\0';
  return charArray;
}

int main( )
{
  char array[ 100 ];  //100 characters in an array
  cout << retvar( array, 100 ) << endl;
}

---

It looks like you were trying to display a variable, retvar, that was out of scope. The variable retvar is specific to the retvar() function only, it can't be used anywhere else. Not only that, you made an integer-returning function that is returning a char variable. What you could do is one of two things:
1)

Code:

---

char * retvar()
{
    char retvar[] = "retvar";
    return retvar;
}

int main()
{
    cout << retvar() << endl;
    return 0;
}

---

OR

2)

Code:

---

char * retvar()
{
    char retvar[] = "retvar";
    return retvar;
}

int main()
{
    char * retvar = retvar();
    cout << retvar << end;
    return 0;
}

---

Brendan

You can do this a bunch of ways, here's one more.

Code:

---

#include <iostream>
#include <cstring>
using namespace std;

char *retvar()
{
        char *ret = new char[sizeof("retvar")];
        return strcpy( ret, "retvar" );
}

int main()
{
        char *string = retvar();
       
        cout<< string <<endl;
        delete string;

        return 0;
}

---

And another that's closer to what you posted but not as handy as other ways.

Code:

---

#include <iostream>
using namespace std;

char *retvar()
{
        return "retvar";
}

int main()
{
        cout<< retvar() <<endl;
        return 0;
}

---

And if you want to return multiple values it's best to either use pointers for your parameters or pass a struct with all of the variables you want to return.

harryP, neither of your suggestions will work because both still try to return a local variable, this variable is destroyed when the function returns and main ends up with garbage. You also can't use the same name retvar for everything or the compiler will panic since it has no idea what you want when you say retvar.

Oops, I was trying to keep it a little like his example whilst getting it to work - guess I'll just give a wroking answer next time :)