Question

First, I read that:

• array
• &array
• &array[0]

will all be the same as long as “array” is really an array.
So I tried:

int main(){ 
     char ar[]={'a','b','c','\0'};
     printf("argument: &ar     %s\n",&ar);
     printf("argument: &ar[1]    %s\n",&ar[1]);
}

the output:

argument:&ar   abc  
argument:&ar[1]  bc

It seems that &ar is taken as a pointer to the first element instead of a pointer to “ar”, which itself is a pointer to the first element if I am not mistaken.

’cause that shouldn’t be how &(pointer to a char) are treated, I tried:

char a='s';
char *pa=&a;
printf("argument: &pa   %c\n",&pa);

The output for %c isn’t even a character.
Nor shouldn’t it be how a pointer to the first element of an array treated. I tried:

char *pa=&ar[0];
char **ppa= &pa;
printf("argument: &pa   %s\n", ppa);

The output for %s is, not surprisingly, nonsense; but why hasn’t &ar been nonsense? For if ar were a pointer, shouldn’t &ar be as much a pointer to a pointer as is ppa?

My questions:

1. Is simply “&” ignored when put before an array name?
   Are array names simply special on this?
2. If so, how does the compiler verify that the identifier following “&” is a reference to an array? Does it actually search for it in a list of arrays so for declared?

Answer 1

Seems that &ar is taken as a pointer to the first element instead of a
pointer to “ar”, which itself is a pointer to the pointer of the first
element if I am not mistaken.

You are mistaken. &ar is a pointer to the array ar, but the array ar is not a pointer of any sort (it’s an array), so &ar is not a pointer to a pointer.

An array is a contiguous sequence of objects – in the case of ar, it’s a contiguous set of 4 chars. &ar is a pointer to this set of 4 chars, which necessarily means it points at the same place as &ar[0], a pointer to the first char in that set. It has a different type, though: &ar has type char (*)[4] which means “pointer to array of 4 chars” and &ar[0] has type char *, which means “pointer to char”.

The confusion arises because in almost all expressions, ar evaluates to a pointer to the first element of the array (the exceptions to this are when it’s the operand of the unary & operator or the sizeof operator). This doesn’t mean that ar is a pointer though – it’s not – just that in most cases it evaluates to a pointer value.