Why does this code snippet not work in C ?

[DelphiCoder]

It says "str" is not initialized.

void SInit(char *s) {
   s = malloc(sizeof(char)*25);
}
 
void SFree(char *s) {
   free(s);
}
 
int main()
{
   char *str;
   SInit(str);
   strcpy(str, "Hello, World");
   printf("%d", strlen(str));
   SFree(str);
   getch();
   return 0;
}

[JosAH]

C (and C++) pass parameters by value, C can't even pass by reference (C++ can).

kind regards,

Jos

[DelphiCoder]

But I'm passing an address. You can do that in C. For example, this works:

void Init(int *n) {
   *n = 5;
}
 
int main()
{
   int *i;
   i = malloc(sizeof(int));
   Init(i);
   printf("%d", *i);
   getch();
   return 0;
}

[JosAH]

@DelphiCoder
You are passing an address here; that function manipulates the object pointed to by that address. That is entiry different from your first example. Think of it: C passes parameters by value; that explains it all.

kind regards,

Jos

[DelphiCoder]

I'm passing an address in the first post too. I understand that C parameters by value.

Let me put it this way: Can you fix the code in the first post so that it works as intended? Maybe then I will understand what you're trying to tell me.

[JosAH]

@DelphiCoder
Sure, here goes:

void SInit(char **s) { 
   *s = malloc(sizeof(char)*25); 
} 
 
void SFree(char *s) { 
   free(s); 
} 
 
int main() 
{ 
   char *str; 
   SInit(&str); 
   strcpy(str, "Hello, World"); 
   printf("%d", strlen(str)); 
   SFree(str); 
   getch(); 
   return 0; 
} 
 

kind regards,

Jos

ps. I would've made the SInit function return the address of the allocated memory.

[weaknessforcats]

What JosAH is saying is that in order to pass a function argument by value, the compiler needs to make a copy of the variable in order to be sure the function does not change the variable used as the argument.

Your first example shows malloc() changing the value of s. This is a copy of the variable used to make the call. In order for the function to change the value of the variable used on the call, you need to pass the address for that variable. That is, pass the address of the pointer.

[DelphiCoder]

I guess there is a gap in my understanding that I wasn't aware of. I don't see why I need to pass a pointer to a pointer.

For instance, given what you guys said, I don't understand why this code works. I'm changing the value inside the function and I don't need a pointer to a pointer to do it. There seems to be something about the "malloc" function that requires a pointer to a pointer, rather than just a pointer. "strcpy", on the other hand, operates on a pointer and the change it makes is done to the original variable, not a copy of it...

void Init(char *s) {
   strcpy(s, "Hello");
}
 
int main()
{
   char *s;
   s = malloc(6 * sizeof(char));
   Init(s);
   printf("%s", s);
   getch();
   return 0;
}

[weaknessforcats]

void SInit(char *s) {
s = malloc(sizeof(char)*25);
}

This was the example.

Here the variable s is changed by the return from malloc().

Then s is destroyed when the function completes.

In order to chane the variable in main(), you need to have the address of that variable. That is, the address of s:

void SInit(char **s) { 
   *s = malloc(sizeof(char)*25); 
} 

Your example using strcpy is exactly the same. That is, to change the character in main(), you need the address of that character. It's the same with a pointer: To change the value of a pointer in main() you need the address of that pointer in your function.

[DelphiCoder]

Thanks a lot. I think I understand now. In most cases, we change the content of the memory block by passing a pointer to it into the function. But what I was trying to do here is change the actual pointer by making it point at a different block of memory, and to do that I have to pass a pointer to that pointer.

Thanks for the explanation.

[weaknessforcats]

I think I understand now. In most cases, we change the content of the memory block by passing a pointer to it into the function. But what I was trying to do here is change the actual pointer by making it point at a different block of memory, and to do that I have to pass a pointer to that pointer.

You do understand that these two cases are identical? Yes?

That is, to change the value at a memory location, you need the address of that location.

[Markus]

@DelphiCoder
*brain fart*

Mark (apologises for being off-topic).

[DelphiCoder]

@weaknessforcats
... and to change the value of the address itself, I need the address of the address.

[JosAH]

@DelphiCoder
Yep, you got it; it is similar to Dephi/Pascal: all the 'var' parameters are passed by reference, the others are passed by value; C can only do the latter.

kind regards,

Jos

[DelphiCoder]

OK, I copied/pasted the code into my VC++ and it doesn't work. I get a 6 digit number instead of the expected string:

void SInit(char **s) { 
   *s = malloc(sizeof(char)*25); 
} 
 
 
int main() 
{ 
   char *str; 
   SInit(&str); 
   strcpy(str, "Hello, World"); 
   printf("%d", str); 
   free(str); 
   getch(); 
   return 0; 
} 

[sridhard2406]

please change the 11 th line as

printf("%s\n",str );

[DelphiCoder]

@sridhard2406
Well duh. I do apologize. I actually spent a half hour struggling with it, never noticing the stupid mistake.