Simple pass-by-reference question

[MimiMi]

How do I get the pass by reference-thing to work in this example?
Since I'm still new to C I sometimes get confused in this matter, and this is one of those occasions.

I'm obviously doing something wrong.

int main(void)
{
char *params =  mem_alloc(sizeof(char)*64);
getParams(params);
printf("Parameters: %s",params);
mem_free(params);
}
 
int getParams(char *p)
{
  p = "something, something";
  return 0;
}

what I want is of course the result to be "Parameters: something, something".
Thanks!

[JosAH]

There doesn't exist a call by reference parameter passing mechanism in C; it only
uses call by value for passing parameters around. You can still copy your string
to that chunk of memory. Change your line 11 as follows:

strcpy(p, "something, something");
 

kind regards,

Jos

[MimiMi]

Ok, thank you very much, my problem is solved!
However, I'm trying to really understand why just writing

p="something,something";
 

doesn't work. While debugging, I see that the params-pointer and the p-pointer both point to the same address. That's why I'm confused.
But is it because the pointer p only gets assigned to point to the string 'locally', so once we're outside the scope of the getParam - method, this assignment is all forgotten about? If that's the case, I think I understand. Anyhow, again thank you so very much!

[JosAH]

But is it because the pointer p only gets assigned to point to the string 'locally', so once we're outside the scope of the getParam - method, this assignment is all forgotten about? If that's the case, I think I understand. Anyhow, again thank you so very much!

Yep, that's how call by value works: just values are passed to a function and
assigned to the parameters as if they were local variables. If you change them
you don't change anything outside that function. If such a value happens to be
an address and you change some memory content at that address, the changes
will be permanent then, i.e. not local to that function.

kind regards,

Jos