a question on int **

[Devil Panther]

I have a question on the next "code", why do I pass to the function func the address of an address? (&var)

int func(int **);

main()
 {
    int *var;
    
    func(&var);
 }

Thanks.

[Dave_Sinkula]

>why do I pass to the function func the address of an address?

http://faq.cprogramming.com/cgi-bin/...1043284351#016

[edit]
It's not "the address of an address", it's "the address of a (pointer) variable".
[/edit]

[quzah]

For a further illustration, look at this neighboring thread.

Quzah.

[Devil Panther]

the address of a pointer? but the address of *var is var

[quzah]

int var;
int *varptr;
int **varptrptr;

var = 5;
varptr = &var;
varptrptr = &varptr;

*varptr gives us 5.
*varptrptr gives us varptr.
**varptrptr gives us 5.

Dereferencing a pointer gives you whatever it points to.

Thus:

*varptr gives you the integer
*varptrptr gives you the pointer.

So **varptr gives you the pointer varptr which in turn is dereferenced, giving you the 5.

Quzah.

[Devil Panther]

so, by & on an address you get it's value?

[quzah]

The & gives you the address of that variable. (IE: Where that variable is in memory.)

Thus:

#include <stdio.h>
int main( void )
{
    int x = 5;
    printf("%p is the address of x.\n", &x );
    printf("%d is the value of x.\n", x );

    return 0;
}

To make a pointer point at something, it either has to be a direct assignment to another pointer, or the address of a variable:

int x;
int *ptr1, *ptr2;

x = 5;

ptr1 = &x; /* make ptr1 hold the address of x */
ptr2 = ptr1; /* make ptr2 point at what ptr1 points at */

Quzah.

[Devil Panther]

I understand!

so &varptr is the address of *varptr, to match varptrptr's address, sort of speak...

[quzah]

You can use parenthesis to illustrate exactly what happens with the double-dereference:

int x;
int *xp;
int *xpp;

x = 5; /* give a value to x */
xp = &x; /* give a value to xp */
xpp = &xp; /* give a value to xpp */

The value of x is 5.
The value of xp is the address of x.
The value of xpp is the address of xp.

*xp /* dereferencing a pointer gives you the value of what it points to, thus 5. */
*xpp; /* dereferencing a pointer gives you the value of what it points to, thus the address of xp. */

**xpp can be expanded visually to:

*(*xpp)

Thus, *xpp is the what xpp points to, which is a pointer. So, dereferencing that pointer gives what that pointer points to, producing 5, which is the value of x.

One more time, "*xpp" gives you the pointer "xp". Dereferencing that pointer gives you x, and the value of 5. You can combine the two dereferences into a single step:

**xpp


Quzah.

[Devil Panther]

I understand!

so &xp is the address of *xp, to match xpp's address, sort of speak...


Also, can i do it directly from: int x to int **x ?


Thanks

[Devil Panther]

But what I don't understand, is if int *xp, **xpp, why: *xpp = xp kills the program? it is support to be the same thing, no?

[quzah]

Try this to clear things up:

#include <stdio.h>
int main( void )
{
    int x, *xp, **xpp;

    x = 5; /* give x a value */
    xp = &x; /* give xp a value, the address of x. */
    xpp = &xp; /* give xpp a value, the address of xp */

    printf("A value is what a variable holds.\n");
    printf("Pointers hold addresses of other variables as their value.\n");
    printf("To get the address of a variable, use '&'\n");
    printf("The value of 'x' is %d.\n", x );
    printf("The address of 'x' is %p.\n", &x );
    printf("The value of 'xp' is %p.\n", xp );
    printf("The address of 'xp' is %p.\n", &xp );

    printf("The value of 'xpp' is %p.\n", xpp );
    printf("The address of 'xpp' is %p.\n\n", &xpp );
    printf("Dereference a pointer to get what it points to.\n");
    printf("To dereference a pointer, use '*'\n");
    printf("Use the list of addresses above to see what is happening when you dereference pointers.\n\n");
    printf("Dereferencing xp gives us: %p\n", *xp );
    printf("Dereferencing xpp gives us: %p\n\n", *xpp );

    return 0;
}

Perhaps that will help clear things up.

[edit] Fixed up %p / %d issues as illustrated below. [/edit]

Quzah.

[Dave_Sinkula]

    printf("The value of 'xp' is %p.\n", xp );
    printf("The value of 'xpp' is %p.\n", xpp );
    printf("Dereferencing xpp gives us: %p\n\n", *xpp );

[quzah]

[edit]
Nevermind, I see what you were pointing out. That's what you ( I ) get for copy/pasting the printf lines.
[/edit]

Quzah.