Multiple returns from one function?

[Ed_Severson]

I need to return three integers from one function to main, but I'm not sure how to specify where to return the variables to main..

What i need to know is how to format the return statement in the function, how to format the function prototype to accept multiple returns, and how to specify where the returned variables are placed in the function call.

I cannot find information on this anywhere. Any help would be greatly appreciated.

[Dave Jay]

A function, by definition, can only return one value.
If you need to retain additional information used in the function, supply pointers in the parameter field or use static/global variables (when appropriate).

[Nutshell]

I think you can return a structure. SO just wrap up the values in a structure variable and return that.

[Shiro]

Or give pass a struct to the function, let and let the function fill the struct with data.

#include <stdio.h>

struct data_s
{
    int data1;
    int data2;
    int data3;
};

void function (struct data_s *data)
{   
    data->data1 = 1;
    data->data2 = 2;
    data->data3 = 3; 
}
            
int main () 
{ 
    struct data_s data;
    
    function (&data);
    
    printf ("%d %d %d\n", data.data1, data.data2, data.data3);

    return 0; 
}

[swoopy]

There are three ways:
1. Use parameters and pass them by reference (& in front of them).
2. Use parameters and make them pointers.
3. Use Nutshell's idea and return a structure of three integers.

Here's an example of method 1.

#include <stdlib.h>
#include <stdio.h>

void calculate(int a, int b, int &sum, int &diff, int &prod, int &quo);
int main(void)
{
   int a = 10;
   int b = 5;
   int sum, diff, prod, quot;

   calculate(a,b,sum,diff,prod,quot);
   printf("%d+%d=%d\n",a,b,sum);
   printf("%d-%d=%d\n",a,b,diff);
   printf("%d*%d=%d\n",a,b,prod);
   printf("%d/%d=%d\n",a,b,quot);
   return 0;
}

void calculate(int a, int b, int &sum, int &diff, int &prod, int &quo)
{
   sum = a + b;
   diff = a - b;
   prod = a * b;
   quo = a / b;
}

[quzah]

Swoopy, you're confused here. You cannot pass by reference in C. That is a C++ feature. This is not valid C code:

void myfun( int &x ) { x = 20; }
int main( void ) {
    int x = 0;
    myfun( x );
    return printf("%d", x ); //x will NOT equal 20
}

This would change the value of x in C++, but it does not in C.

Quzah.

[Led Zeppelin]

You may want to review formal and actual parameters. You use 1 function and multiple function calls. I am still new myself but I believe this is a viable solution. Shiro gives a good example of what I am trying to say. Good Luck.

[swoopy]

Thanks Quzah. I did not know c did not have this feature.

[swoopy]

I'll repost my example with pointers, since references aren't c.

#include <stdlib.h>
#include <stdio.h>

void calculate(int a, int b, int *sum, int *diff, int *prod, int *quo);
int main(void)
{
   int a = 10;
   int b = 5;
   int sum, diff, prod, quo;

   calculate(a,b,&sum,&diff,&prod,&quo);
   printf("%d+%d=%d\n",a,b,sum);
   printf("%d-%d=%d\n",a,b,diff);
   printf("%d*%d=%d\n",a,b,prod);
   printf("%d/%d=%d\n",a,b,quo);
   return 0;
}

void calculate(int a, int b, int *sum, int *diff, int *prod, int *quo)
{
   *sum = a + b;
   *diff = a - b;
   *prod = a * b;
   *quo = a / b;
}