Why (and when) do I need to use parentheses after sizeof?

Question

The below fails to compile:

typedef int arr[10]; int main(void) {     return sizeof arr; }  sizeof.c:3: error: expected expression before ‘arr’ 

but if I change it to

sizeof(arr); 

everything is fine. Why?

Answer 1

According to 6.5.3, there are two forms for sizeof as the following:

sizeof unary-expression sizeof ( type-name ) 

Since arr in your code is a type-name, it has to be parenthesized.

Answer 2

That's the way the language is specified, type names must be parenthesized here.

Suppose the grammar looked like this:

sizeof unary-expression sizeof type-name

Now, e.g. the following expression would be ambiguous:

sizeof int * + 0 

It could be either sizeof(int *) + 0 or sizeof(int) * +0. This ambiguity doesn't arise for unary expressions, as an asterisk appended to an expression isn't an expression (but for some type names, appending one, is again a type name).

Something had to be specified here and requiring type-names to be parenthesized is a way to solve the ambiguity.