Thread: Call-by-reference

If I have a simple program like this:

Code:

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

void change (char *t);

int main (void)
{
char *test = "hellooo\n";
printf("%s", test);
change(test);
printf("%s", test);
return 0;
}

void change (char *t)
{
t = "Hi!\n";
}

How do I make it so that test would become "Hi!"? What am I doing wrong? THanks.

Either return the pointer, or make the function parameter a pointer to a pointer.

Code:

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

void change (char **t);

int main (void)
{
char *test = "hellooo\n";
printf("%s", test);
change(&test);
printf("%s", test);
return 0;
}

void change (char **t)
{
*t = "Hi!\n";
}

Sorenson,

That was useful! By returning the pointer do you mean something like:

Code:

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

char *change (){
  return "Hi!\n";
}

int main (void){
  char *test = "hellooo\n";
  printf("%s", test);
  printf("%s", change(test));
  return 0;
}

The above works but I have had difficult experinces with strings and welcome your comments! Incidentally, being lazy and wishing to avoid errors I always put my main last - is there any benefit in doing an initial declaration of the function - it seems needlessly repetitive?

>By returning the pointer do you mean something like ..snip

Yep.


>is there any benefit in doing an initial declaration of the function - it seems needlessly repetitive?

I don't think so, but if you've a lot of functions it'll be easier to check their parameters, and return type if you've a list of prototypes at the top, and they could be in different files.

char *test = "hellooo\n";
//I think if you are going to change "test", if needs to be declared:
char test[30] = "hellooo\n";
//Although what you have will probably work.

void change (char *t)
{
strcpy(t,"Hi!\n");
}