C function pointer syntax

user1941583 picture user1941583 · Jan 1, 2013 · Viewed 36.3k times · Source

My question is a very simple one.

Normally, when declaring some variable, you put its type before it, like:

int a;

a function pointer may have type like: int(*)(int,int), in case we point to a function that takes two integers and returns an integer. But, when declaring such a pointer, its identifier is not after the type, like:

int(*)(int,int) mypointer;

instead, you must write the identifier in the middle:

int(*mypointer)(int,int);

why is this so? Sorry, I know it's an embarrassingly easy question...

Thanks to everybody for replying. A.S.

Answer

detly picture detly · Jan 1, 2013

I explain this in my answer to Why was the C syntax for arrays, pointers, and functions designed this way?, and it basically comes down to:

the language authors preferred to make the syntax variable-centric rather than type-centric. That is, they wanted a programmer to look at the declaration and think "if I write the expression *func(arg), that'll result in an int; if I write *arg[N] I'll have a float" rather than "func must be a pointer to a function taking this and returning that".

The C entry on Wikipedia claims that:

Ritchie's idea was to declare identifiers in contexts resembling their use: "declaration reflects use".

...citing p122 of K&R2.