Semantics of char a[]

Question

I recently embarrassed myself while explaining to a colleague why

char a[100];
scanf(\"%s\", &a); // notice a & in front of \'a\'

is ve

Answer 1

It's been a while since I programmed in C but here's my 2c:

char a[100] doesn't allocate a separate variable for the address of the array, so the memory allocation looks like this:

 ---+-----+---
 ...|0..99|...
 ---+-----+---
    ^
    a == &a

For comparison, if the array was malloc'd then there is a separate variable for the pointer, and a != &a.

char *a;
a = malloc(100);

In this case the memory looks like this:

 ---+---+---+-----+---
 ...| a |...|0..99|...
 ---+---+---+-----+---
    ^       ^
    &a  !=  a

K&R 2nd Ed. p.99 describes it fairly well:

The correspondence between indexing and pointer arithmetic is very close. By definition, the value of a variable or expression of type array is the address of element zero of the array. Thus after the assignment pa=&a[0]; pa and a have identical values. Since the name of the array is a synonym for the location of the initial element, the assignment pa=&a[0] can also be written as pa=a;