s[ ] is an array.&s should give address of decayed pointer

Question

After initialising character array name and address refers to same. s[] is an array. so it decays to pointer s, &s should give the address of pointer s

#include <stdio.h>
int main()
{
   char s[]="Get Organised! learn C!";
   printf("%u %s\n",&s[2],&s[2] );
   printf("%u %s\n",s,s );  /* This line and below line is equivalent*/
   printf("%u %s\n",&s,&s ); /* Please Explain */
   printf("%u %s\n",s[2],s[2] );
   printf("%u %s\n",s[3],s[3] );
   return 0;
}

%u is just used to see what is going inside.

It's all undefined behaviour. `%u` requires that you're passing an `int` or `unsigned int`, and you don't. — Kerrek SB, Mar 10 '16 at 19:07
No, I am asking **printf("%s\n",s );** **printf( %s\n",&s );** gives same answer — Saket Anand, Mar 10 '16 at 19:09
s[ ] is an array. so it decays to pointer **s** , **&s** should give the address of pointer **s** — Saket Anand, Mar 10 '16 at 19:14
Ack, you changed the question on me. Short answer - `s` is *not* converted to a pointer when it is the operand of the unary `&` or `sizeof` operators, so the *values* of `s` and `&s` are the same (address of the first element of the array), but the *types* are different (`char *` and `char (*)[28]`, respectively). — John Bode, Mar 10 '16 at 19:19
@JohnBode : - I copied your answer. I could not understand that well so i put in notepad so that i can read it later.Thank You very Much — Saket Anand, Mar 10 '16 at 19:37
Summary 'C is inconsistent and borderline insane when it comes to array handling'. Any sane language would have forced to to pass '&array' to pass by address but, unfortunately, C was written by Unix developers who needed to handle text as easily as possible, (text is all they understand), so no '&'. — Martin James, Mar 10 '16 at 19:59
Even if `&s` did give you the address of the pointer it decays to, that wouldn't make sense, because that pointer is *not* a variable. (It would be like saying `&1` to get the address of `1`) — user253751, Mar 10 '16 at 20:16

score 1 · Answer 1 · answered Mar 10 '16 at 19:19

Getting rid of undefined behaviour first:

int main() {
  char const string [] = "hello";
  printf("%p %p\n", (void *) string, (void *) &string);
  return 0;
}

You get (probably, that's target dependent) the same address printed here because the first expression is implicitly converted to a pointer to the first element of the array while the second is a pointer to the whole array (which most likely happens to start at its first element)

string == &(string[0]) // this is often called decaying

The type of the expression string (or more precisely the expression it decays to) is char const * whereas the type of &string is char const (*array)[6], so they're not "the same".

The reason that &string is not a pointer to a pointer is simple: The C standard explicitly forbids array expressions prefixed with the address of operator (&) (or inside an sizeof operator) to be implicitly converted to a pointer.

score 0 · Accepted Answer · edited May 23 '17 at 12:31

With the below, s is an array 100 of char. When is is passed to printf(), array s is converted to the value and type of the address of first element. So printf() is given a char *.

&s is the address of an array 100 of char. It has an equivalent value as s, yet a different type.

char s[100]="Whatever";
printf(some_format,s);
printf(some_format,&s);

To properly print addresses of variables, literals, cast to (void*) and use "%p". The print out typically lacks any description of the type of the variable.

printf("%p\n",(void*) s);
printf("%p\n",(void*) &s);

Output

0x28cbdc
0x28cbdc

To see an effect of the type difference:

printf("%zu\n",sizeof *(s));
printf("%zu\n",sizeof *(&s));

Output

1
100

@john bode also suggested this earlier.

s[ ] is an array.&s should give address of decayed pointer

2 Answers2