C memory location of the array name

Question

I'm confused about how C stores in memory an array. From the example bellow the address of str1 is equal to its value. Whit this I was interpreting the result as: the memory address &str1 contains the value of the address itself. But this can not be the case because from other examples I saw the address of str1 contains the first value of the string, which in this case is "H".

So my question can be: where is the address of str1 stored?

#include <stdio.h>

int main(int argc, char const *argv[]) {
  char str1[] = "Hello World";
  printf("%d %d %c\n",&str1,str1,str1[0] );
  return 0;
}

When you say "address of `str1`", what do you exactly mean by that? — Eugene Sh., Oct 17 '17 at 13:37
`str1` is an array so using that name as you do here results in its address. `&str1` evaluates to the same value as `str1`. `str1[0]` is the vaue of its first element. — Paul Ogilvie, Oct 17 '17 at 13:39
@PaulOgilvie `&str1` and `str1` evaluate to the same value but different types. — Eugene Sh., Oct 17 '17 at 13:40
@EugeneSh, correct, but not relevant here. (Updated my comment) — Paul Ogilvie, Oct 17 '17 at 13:41
For `int b;` , where is `&b` stored? Once you can answer this question, it is in fact the same answer as for `&a`. — M.M, Oct 17 '17 at 15:26

Sourav Ghosh · Answer 1 · 2017-10-17T13:43:16.930

First of all, you must use %p format specifier to print an address (pointer).

That being said, quoting C11, chapter §6.3.2.1

Except when it is the operand of the sizeof operator, the _Alignof operator, or the unary & operator, or is a string literal used to initialize an array, an expression that has type ‘‘array of type’’ is converted to an expression with type ‘‘pointer to type’’ that points to the initial element of the array object and is not an lvalue.

So, basically,

  printf("%p", str1);
  printf("%p", &str1[0]);

are the same.

Now, coming to the part of &str1, well, the type is different. It is of type char [<array size>].

C memory location of the array name

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