Hi everyone...

    I'm a student of C and I'm confused about this question involving
pointers.

    In the following code I'm confused about what d points to. Here's my
(obviously wrong) thinking...

b is a pointer to char which points to the 4th char of array a[]... ('t');

c is a pointer to a pointer to char which is assigned the address
 of b...so it holds the address of b. So c points to b.

Then we come to my problem:  *d = *c. It seems to say that d (which
is a pointer to char) is assigned the dereferenced value of c. In
other words, it would seem that d and c both point to b.

But running the program prints "Abbey" which means that d points to
't' and d+5 is the string "Abbey" which means that d doesn't point to
b but points to the SAME address that b does.

HELP!! Thanks...Stephen.  ;>)

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

main()
{
 char a[] = "Tintern Abbey";
 char *b=a+3;
 char **c=&b;
 char *d=*c;
 puts(d +5);

--

     *
   *-|-*      Stephen Levinson

   *-|-*
     *

--

Here's the problem; since c points to b, *c points to what b points to;
the declaration/initializer syntax should be thought of as char *d
and then d = *c.  (I misread this in a post a week or so ago when
there were two variables declared in the same line and I read the
comma as a semicolon.)  Notice that d has the wrong type to point
to b; it's char *, and c has type char **.

Exactly; this is the result of d = *c.

This acts like char *d; d = *c;

--
MJSR

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Ah, this is what happens when people get too cute with initializers.
Here is an (almost) equivalent version:

int main(void) {

        char a[] = "Tintern Abbey";
        char *b, **c, *d;

        b = a + 3;      /* 'b' points to the letter 't' */
        c = &b;             /* 'c' points at the variable 'b' */
        d = *c;         /* 'd' now has the value of the pointer 'c'
                            points at, ie., what 'b' points at */
        puts( d + 5 )   /* or a + 8 */

Regards,

Anton

--

Be careful here.  All these initialization of b, c, d are to non-constants.
This is a dangerous road to start down.

Let's look at a modified form of your code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(void)
{
    char a[] = "Tintern Abbey";
    char *b = a + 3;
    char **c = &b;
    char *d = *c;
    printf("a is %p; the string is %s\n", (void *)a, a);
    printf("b is %p; the string is %s\n", (void *)b, b);
    printf("c is %p; *c is %p; the string is %s\n",
            (void *)c, (void *)*c, *c);
    printf("d is %p; the string is %s\n", (void *)d, d);
    printf("d+5 is %p; the string is %s\n", (void *)(d+5), d+5);
    puts(d + 5);
    return 0;

The values of the pointers will obviously be different for you, but they are
useful for explication.  I get the following.

a is 2e0fd0; the string is Tintern Abbey
b is 2e0fd3; the string is tern Abbey
c is 2e0fcc; *c is 2e0fd3; the string is tern Abbey
d is 2e0fd3; the string is tern Abbey
d+5 is 2e0fd8; the string is Abbey
Abbey

 ==

OK:
a is 2e0fd0; the string is Tintern Abbey
b is 2e0fd3; the string is tern Abbey

 ==

  ==

OK:
b is 2e0fd3; the string is tern Abbey
c is 2e0fcc; *c is 2e0fd3; the string is tern Abbey

  ==

  ==

No ...
b is 2e0fd3; the string is tern Abbey
c is 2e0fcc; *c is 2e0fd3; the string is tern Abbey
d is 2e0fd3; the string is tern Abbey

c points to b; d is a pointer with the same value as b (*c).

  ==

--


--

Groovy hepcat Stephen Levinson was jivin' on Mon, 24 May 1999 21:25:08
GMT in comp.lang.c.
Pointers to Pointers to Char in C's a cool scene! Dig it!

  Yes, that's right.

  Also correct.

  Yes. That means d is assigned the value of what c points to, which
is b.

  No, no. d points to what b points to, since d equals b. It's like
this: c points to b, therefore *c (dereferenced c, or the thing c
points to) is b. Since d is initialised with *c, which is b, it's the
same as saying

char *d = b;

  Yep.

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This could be rewritten equivalently

char a[] = "Tintern Abbey";
puts(&a[8]);

or, if you prefer

puts((a + 3) + 5);

since this is what is really happening.
--