Question

for some reason the code I am looking at has a lot of something like


char tmp[4];

memcpy(&tmp[0],foo_pointer,bar_size)


I would have expected simply


char tmp[4];

memcpy(tmp,foo_pointer,bar_size)


Is there some reason I am missing in writing it the first way?

Answer 1

No, they’re equal.

The former might be clearer even without the definition of tmp in sight.

(And just to make this answer complete as to why it is so, here’s the interesting bit from a late draft of the standard:)

(6.3.2.1p3) 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. If the array object has
register storage class, the behavior is undefined.