C passing a global pointer

[aaa bbb]

Hi guys.
Let's say I've got something like this:

struct Box{
//something
};
typedef struct Box* pBox;

void fun(pBox p);

int main(){
pBox p = NULL;
fun(p);
}

void fun(pBox p){
if(p == NULL) p = (pBox)malloc(sizeof(Box));
}

function fun recieves the address(which is NULL) and then allocates the memory for the Box; Let's say I cannot return the address of new allocated p and I can't also use that pointer p(from main) without passing it into a function.

Q: How can I make it that I could operate in function "fun" as I operate on orginal pointer p(from main), right now I'm just passing the address to my function but I can't change
the 'global' pointer p ;(.

I remember in pascal it's like:
"function(var pointer:[pointer to sth])"
and all is done.

How can I do it in C?
Any ideas?

[johny10151981]

Read About IPC. there is lots of way to do this.

[weaknessforcats]

This function:

void fun(pBox p){
etc...

makes a copy of the pointer used by the caller. What you want to do is tell the compiler to not make that copy. In C++ that is called a reference:

void fun(pBox& p){
etc...

Now in fun, the argument p is the same pointer used by the calling function. No copy was made.

[aaa bbb]

I tried "pBox& p" but it says:
'expected ';', ',', ')' before '&' token.
I'm nut sure if I can do that,
isn't it like "Box* &p"?...

Guess it's not available in C. So it's not possible to use in function 'original' pointer?(not using global pointer in functions, gotta pass them first)

[weaknessforcats]

Did you change both the function and the prototype?

Here is your code with the change and it compiles OK:

struct Box{
 //something
 };
 typedef struct Box* pBox;
 
void fun(pBox& p);
 
int main(){
 pBox p = NULL;
 fun(p);
 }
 
void fun(pBox& p){
 if(p == NULL) p = (pBox)malloc(sizeof(Box));
 }

[Banfa]

void fun(pBox& p)

This doesn't work because it is a reference which is only supported by C++, a solution in C is required. And that is simple, just use a pointer to a pointer.

Whenever you want to output some type but not use the return value you need a pointer to that type in your function prototype, for int

int main()
{
  int a;
 
  getValue(&a);
 
  printf("The value is: %d\n", a);
 
  return 0;
}
 
void getValue(int* pValue)
{
  *pValue = 10;
}
 

This is true for any type and can generalised for any type A as

void getValue(A* pValue)
{
  *pValue = <ValidValueForTypeA>;
}
 

In your case you want to pass out a variable of type pBox so in the code above A = pBox and you need this prototype void fun(pBox* p).

In full your function becomes

void fun(pBox* p)
{
  if(*p == NULL)
  {
    *p = malloc(sizeof(Box));
  }
}
 

Note the is no need and indeed it can cause subtle errors if you cast the return value of malloc in C.

[weaknessforcats]

fun(pBox* p)
 2. {
 3.   if(*p == NULL)
 4.   {
 5.     *p = malloc(sizeof(Box));
 6.   }
 7. }
 

This won't work because p is a Box* so *p s a Box.

I believe you want:

fun(pBox** p)
  {
   if(*p == NULL)
    {
      *p = malloc(sizeof(Box));
   }
  }
 

Now p is a pointer to a Box* so *p is now a Box*.

[Banfa]

This won't work because p is a Box* so *p s a Box.

No p is pBox* and pBox is typedef'd to Box*, making p a Box**, just what is wanted. If you use pBox** p then p would be Box*** which would be indirection gone mad :D

[aaa bbb]

why does it give me runtime error when I try go with '*p == NULL'?

[Banfa]

You probably call the function with the wrong variable type, you are trying to write a variable of type pBox so you need to declare a pBox in main.

[aaa bbb]

okay, what I do wrong?:

typedef struct {
 int n;
}Box;
typedef struct Box* pBox;
 
void fun(pBox* p)
{
  if(*p == NULL)
  {
    *p = malloc(sizeof(Box));
  }
}
 int main(){
    pBox p = NULL;
    fun(p);
}

still got that error in '*p == NULL', can someone compile this and check, it kills me...

[weaknessforcats]

You need to #include <stdio.h> in order to get NULL declared.

You may have other warnings which indicate a weakness in the code. Ideally, you should have no errors and no warnings.

[aaa bbb]

yes, I included libs but the problem comes when I run the program.
It appears runtime error and points the line with NULL. But I guess it's sth wrong with the *p == null because when I change it to 'p' it works.

[Banfa]

Do you not get any warnings when you compile? I get the following warning compiling your code

bytes.c:15: warning: passing argument 1 of 'fun' from incompatible pointer type

You should not ignore compiler warnings, they are there to tell you when you are doing something suspect. This is one of those cases where C++ is stricter than C it would not allow that line to compile, it produces an error rather than a warning. However that is correct the compiler assumes you know what you are doing and allow the pointer conversion to go ahead.

So you pass p of type Box* to a function expecting Box**. p has the value NULL or 0 and the first thing you do is dereference it at line 8 since it is NULL the program crashes.

The warning highlights the problem, you passed type Box* to a function expecting Box**, you passed the wrong thing, you needed to dereference the pointer you were passing as the example in my first post when you want to output an int the function takes int* and you pass &a (where a is an int variable). The function is expecting pBox* so you pass &p (where p is a pBox variable).

i.e. line 15 should be fun(&p);

Don't ignore compiler warnings and if you didn't get a warning get a better compiler.

[aaa bbb]

Oh, now I see... should have read your previous reply better. And thanks for advice with the warnings, I simply didn't care what they said thinking it's not the case of my problem.
Thanks for help. ;D