Passing variables to functions

**theitsmith** · 02-06-2012

Hi,

I'm trying to pass a variable to testVars(), have testVars() change the variable, and then make that variable available to main().

Passing the variable from main() to testVars() is working--getting it back to main() is not.

Code:

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

char testVars(char *a, char *b);

main()
{
    char * testChar = "We just passed";
    char * testCharTwo = "a main() variable!";
    testVars(testChar, testCharTwo);

    printf("%s %s - Back to main()!\n", testChar, testCharTwo);

    return 0;
}

char testVars(char *a, char *b)
{
    b = "a variable!";
    printf("%s %s\n", a, b);
}

The expected result is that the program will print the following two lines:

We just passed a variable!
We just passed a variable! - Back to main()!

But what I'm getting is:

We just passed a variable!
We just passed a main() variable! - Back to main()!

I'm new to C and am still in the early stages of learning it, but it's my understanding that if I pass a pointer variable from main(), in my case, to testVars() that testVars() is actually working with the dereferenced location of testCharTwo (since this is the only one I wish to change in testVars()) and any modifications it makes to testCharTwo should be immediately available to main().

Am I missing something here? Or is my understanding of this not correct?

~ Tom

**hk_mp5kpdw** · 02-06-2012

Within your function testVars, the variable b is local to the function and changes to it disappear once the function ends. This is the same as when passing anything to a function, if you need to have it changed by that function when it ends, you need to pass a pointer. In this case what you are passing in is already a pointer to begin with but to have that changed by the function what you need to pass is still the address of that variable (a pointer to a pointer).

Code:

void test(const char** testp);

int main()
{
    const char * mytestptr = "Hello World!\n";

    printf("%s",mytestptr);
    test(&mytestptr);  // Pass address of mytestptr, a pointer to a pointer
    printf("%s",mytestptr);

    return 0;
}

void test(const char** testp)
{
    *testp = "Goodbye World\n";
}

**giove** · 02-06-2012

You are working in C.
b is a char *.
In the function you assign to this "a variable!" this means that the address of this string (that is somewhere in the memory) is used in testVars.
When the function returns, b points again the initial address.
You can try with a strcpy(b,"a variable!"); this copies the content in the actual address b. But be sure that the new string is shorter than the first!

**kevinstrijbos** · 02-06-2012

You need to use the dereference symbol at b = "a variable"

**theitsmith** · 02-06-2012

Originally Posted by hk_mp5kpdw

Within your function testVars, the variable b is local to the function and changes to it disappear once the function ends. This is the same as when passing anything to a function, if you need to have it changed by that function when it ends, you need to pass a pointer. In this case what you are passing in is already a pointer to begin with but to have that changed by the function what you need to pass is still the address of that variable (a pointer to a pointer).

Thanks, this worked!

Code:

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

void testVars(char *a, char** b);

int main()
{
    char * testChar = "We just passed";
    char * testCharTwo = "a main() variable!";
    testVars(testChar, &testCharTwo);

    printf("%s %s - Back to main()!\n", testChar, testCharTwo);

    return 0;
}

void testVars(char *a, char** b)
{
    *b = "a variable!";
    printf("%s %s\n", a, *b);
}

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