typo in function pointer variable declaration, but

问题:

While answering a question about function pointers, OP did that to declare a function pointer which takes 1 integer argument and returns nothing:

void *(intr_handlerptr)(int);  // wrong but compiles!!
intr_handlerptr = intr_handler;  // compiler error: cannot assign to this weird thing (lvalue required as left operand of assignment)

when the proper declaration is

void (*intr_handlerptr)(int);  // correct

What's funny is that the error occurs when assigning to this function pointer, not when declaring it (tested with gcc 7.3.1)

So what does that first line do?

回答1:

The expression void *(intr_handlerptr)(int); reads as "intr_handlerptr is a function taking an int and and returning a pointer to an unnamed thing (a void pointer). So the parenthesis around the function name have no use and are discarded. This is a prototype.

The expression void (*intr_handlerptr)(int); reads as "intr_handlerptr is a pointer to a function taking an int and returning nothing. The parenthesis are necessary to associate the name with the pointer type. This is a variable.

When assigning to the first, the compiler complains that you try to assign to "nothing" because the left-hand symbol is not a variable. Actually, the compiler only knows a prototype of that name in its symbol table. Of course, the second assignment statement is correct because you are assigning to a variable.