Basic pointer scope Q

jnusa · 11-19-2004, 05:01 AM

I´m having problems with pointers within pointer scope. In the following code. See following code:

Code:

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

void another_func(char *ptr)
{
ptr = malloc(4);
*(ptr+3) ='\0';
strcpy(ptr, "Hej");
printf("\nptr: %s", ptr);
}

void some_func(char *ptr)
{
another_func(ptr);
printf("\nptr: %s", ptr);
}

int main (int argc, char **argv)
{
char *ptr;
some_func(ptr);
printf("\nptr: %s", ptr);
return(0);
}

The printout of this code is:

Code:

Ptr: Hej
Ptr: 
Ptr:

Why doesn't it print in the other 2 function scopes (function prototype should remain the same)?

ta0kira · 11-19-2004, 05:18 AM

When you call 'another_func', the pointer 'ptr' from 'some_func' is passed by value. This means that the local variable 'ptr' is a copy of the value of the 'ptr' that was passed to it from 'some_func'. When you call 'malloc', you are reassigning the local version of 'ptr' to a new address, but since it is a copy, the 'ptr' in 'some_func'/'main' that was passed to it remains unchanged. Try changing the argument type of 'ptr' in your functions to 'char*&' instead of just 'char*'.
ta0kira

[EDIT]
Sorry, I was a little unclear. The addition of '&' makes the argument a reference to a pointer. This means that the value of the 'ptr' varable passed to the functions can be changed. It is just like having your argument type 'char**', except that you don't have to dereference it to change the value. Also, you need to use the 'free' function somewhere in there to deallocate the memory allocated by malloc.
ta0kira

Code:

void another_func(char **ptr)
{
	*ptr = malloc(4); //<- reassigns the pointer passed to another_func
	//...
}

// ----->

void another_func(char *&ptr)
{
	ptr = malloc(4);  //<- reassigns the variable passed to another_func
	//...
}

jnusa · 11-19-2004, 05:27 AM

Hmm... well I thougth that I was passeing the address of the pointer ptr, which was assigned, when i declared the var in main. If I try to change the function prototype to

Code:

 
void another_func(char *&ptr);

I get a compiler error ["syntax error before & token"]
Any other suggestions?

LauroMoura · 11-19-2004, 05:34 AM

Declare:

Code:

void another_func(char **ptr)

Call:

Code:

another_func(&ptr)

Do this for both functions.

ta0kira · 11-19-2004, 05:35 AM

That's right, I forgot it was just C...
ta0kira

jnusa · 11-19-2004, 06:42 AM

I can make the void some_func(char **ptr) work, and just dereference when in use. I still can't get the void another_func(char *&ptr) (I assume you mean '*&' within the function prototype right?) syntax to work. The compiler gives the same error. The *& seems to be the best solution, because you don't have to dereference the variable every time. Could you alter the eksisting code, with the '*&' example, so I can see prototypes and function calls. Thanks

[EDIT]

Didn't see last 2 post. Makes sense then. The '*&' is for C++ then?

/Jnusa

ta0kira · 11-20-2004, 01:47 AM

Yes, that is C++. Sorry.
ta0kira

jnusa · 11-22-2004, 07:23 AM

Does the ** pointer notation not apply to structs (would think of a char as a struct too, I would think)? I've tried to do a similar test code, because I wanted to pass all my var's this way, but it doesn't seem to recognize my parameter as a pointer to my struct.
Test code:

Code:

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


typedef struct{
        char *ptr;
}obj1;

void another_func(obj1 **ptr)
{                
                
        *ptr->ptr = malloc(4);
        *(ptr->ptr+3) ='\0';
        strcpy(*ptr->ptr, "Hey");
        printf("\nptr: %s", *ptr->ptr);
}

void some_func(obj1 **ptr)
{
        another_func(ptr);
        printf("\nptr: %s", *ptr->ptr);
}

int main (int argc, char **argv)
{
        obj1 obj;
        obj1 *ptr;
        ptr = &obj;
        some_func(&ptr);   //or '&obj' ?
        printf("\nptr: %s", ptr->ptr);
        return(0);
}

Am I doing this all wrong?

itsme86 · 11-22-2004, 09:14 AM

Operator precedence has got you here:
your code:

Code:

void some_func(obj1 **ptr)
{
        another_func(ptr);
        printf("\nptr: %s", *ptr->ptr);
}

*ptr->ptr takes the member named ptr of the struct ptr and then dereferences it. What you need to do instead is:

Code:

        printf("\nptr: %s", (*ptr)->ptr);

The same theory goes for another_func().