pass strings by reference in C

Question

I'm having trouble figuring out how to pass strings back through the parameters of a function. I'm new to programming, so I imagine this this probably a beginner question. Any help you could give would be most appreciated. This code seg faults, and I'm not sure why, but I'm providing my code to show what I have so far.

I have made this a community wiki, so feel free to edit.

P.S. This is not homework.

This is the original version

#include <stdio.h>

#include <stdlib.h>
#include <string.h>

void
fn(char *baz, char *foo, char *bar)
{
     char *pch;

     /* this is the part I'm having trouble with */

     pch = strtok (baz, ":");
     foo = malloc(strlen(pch));
     strcpy(foo, pch);

     pch = strtok (NULL, ":");
     bar = malloc(strlen(pch));
     strcpy(bar, pch);

     return;
}

int
main(void)
{
     char *mybaz, *myfoo, *mybar;

     mybaz = "hello:world";

     fn(mybaz, myfoo, mybar);

     fprintf(stderr, "%s %s", myfoo, mybar);
}

UPDATE Here's an updated version with some of the suggestions implemented:

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

#define MAXLINE         1024

void
fn(char *baz, char **foo, char **bar)
{
     char line[MAXLINE];
     char *pch;

     strcpy(line, baz);

     pch = strtok (line, ":");
     *foo = (char *)malloc(strlen(pch)+1);
     (*foo)[strlen(pch)] = '\n';
     strcpy(*foo, pch);

     pch = strtok (NULL, ":");
     *bar = (char *)malloc(strlen(pch)+1);
     (*bar)[strlen(pch)] = '\n';
     strcpy(*bar, pch);

     return;
}

int
main(void)
{
     char *mybaz, *myfoo, *mybar;

     mybaz = "hello:world";

     fn(mybaz, &myfoo, &mybar);

     fprintf(stderr, "%s %s", myfoo, mybar);

     free(myfoo);
     free(mybar);
}

Answer 1

the code most likely segfaults because you are allocating space for the string but forgetting that a string has an extra byte on the end, the null terminator.

Also you are only passing a pointer in. Since a pointer is a 32-bit value (on a 32-bit machine) you are simply passing the value of the unitialised pointer into "fn". In the same way you wouldn't expact an integer passed into a function to be returned to the calling function (without explicitly returning it) you can't expect a pointer to do the same. So the new pointer values are never returned back to the main function. Usually you do this by passing a pointer to a pointer in C.

Also don't forget to free dynamically allocated memory!!

void
fn(char *baz, char **foo, char **bar)
{
     char *pch;

     /* this is the part I'm having trouble with */

     pch = strtok (baz, ":");
     *foo = malloc(strlen(pch) + 1);
     strcpy(*foo, pch);

     pch = strtok (NULL, ":");
     *bar = malloc(strlen(pch) + 1);
     strcpy(*bar, pch);

     return;
}

int
main(void)
{
     char *mybaz, *myfoo, *mybar;

     mybaz = "hello:world";

     fn(mybaz, &myfoo, &mybar);

     fprintf(stderr, "%s %s", myfoo, mybar);

     free( myFoo );
     free( myBar );
}

Answer 2

Other answers describe how to fix your answer to work, but an easy way to accomplish what you mean to do is strdup(), which allocates new memory of the appropriate size and copies the correct characters in.

Still need to fix the business with char* vs char**, though. There's just no way around that.