alcuadrado alcuadrado - 1 year ago 113
C Question

Printing pointers in C

I was trying to understand something with pointers, so I wrote this code:

#include <stdio.h>

int main(void)
char s[] = "asd";
char **p = &s;

printf("The value of s is: %p\n", s);
printf("The direction of s is: %p\n", &s);

printf("The value of p is: %p\n", p);
printf("The direction of p is: %p\n", &p);

printf("The direction of s[0] is: %p\n", &s[0]);
printf("The direction of s[1] is: %p\n", &s[1]);
printf("The direction of s[2] is: %p\n", &s[2]);

return 0;

When compiling it with gcc I get these warnings:

$ gcc main.c -o main-bin -ansi -pedantic -Wall -lm
main.c: In function ‘main’:
main.c:6: warning: initialization from incompatible pointer type
main.c:9: warning: format ‘%p’ expects type ‘void *’, but argument 2 has type ‘char (*)[4]’
main.c:11: warning: format ‘%p’ expects type ‘void *’, but argument 2 has type ‘char **’
main.c:12: warning: format ‘%p’ expects type ‘void *’, but argument 2 has type ‘char ***’

(The flags for gcc are because I must be C89)

Why incompatible types of pointer? Isn't the name of an array a pointer to it's first element? So if s is a pointer to 'a', &s must be a char **, no?
And why do I get the other warnings? Do I have to cast the pointers with (void *) in order to print them?

And when running I get something like this:

$ ./main-bin
The value of s is: 0xbfb7c860
The direction of s is: 0xbfb7c860
The value of p is: 0xbfb7c860
The direction of p is: 0xbfb7c85c
The direction of s[0] is: 0xbfb7c860
The direction of s[1] is: 0xbfb7c861
The direction of s[2] is: 0xbfb7c862

How can the value of s and it's direction (and of course the value of p) be the same?

Thank you very much,


Answer Source

"s" is not a "char*", it's a "char[4]". And so, "&s" is not a "char**", but actually "a pointer to an array of 4 characater". Your compiler may treat "&s" as if you had written "&s[0]", which is roughly the same thing, but is a "char*".

When you write "char** p = &s;" you are trying to say "I want p to be set to the address of the thing which currently points to "asd". But currently there is nothing which points to "asd". There is just an array which holds "asd";

char s[] = "asd";
char *p = &s[0];  // alternately you could use the shorthand char*p = s;
char **pp = &p;
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