On Tue, Dec 10, 2002 at 09:40:36PM -0800, Brendan Moran wrote:
> On most C syntax I do just fine, but function pointers are always a
> head-scratcher for me.
>
> If I wanted to declare a pointer to this function:
>
> int foo(int bar);
>
> What would be the syntax for the declaration?
>
> The best I came up with was this:
>
> (int)(*)(int) pfoo = &foo;
>
> but that doesn't compile.

It's actually very simple once you realize that the function () have the
highest precedence. Couple that with the fact that you want to define a
function pointer, and you're in business.

Start be defining all of the elements:

int *pfoo(int) = foo; // Note that the & is redundant in front of foo

This won't compile but is the correct format. The reason is precedence (i.e.
5+3*2 is 11 not 16). pfoo binds to the function () first so pfoo is a function
that returns an int pointer.

So to change the precedence add parenthesis. pfoo is
first an foremost a pointer, so put parens around the * and pfoo so that it
binds first.

int (*pfoo)(int) = foo; // Works every time!

It's just that simple. pfoo must now be a pointer above all else and since
it has function parens following it's now a pointer to a function.

BAJ

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