What do the parentheses around a function name mean?

好久不见. 提交于 2019-11-26 23:34:02
NPE

In the absence of any preprocessor stuff going on, foo's signature is equivalent to

int foo (int *bar)

The only context in which I've seen people putting seemingly unnecessary parentheses around function names is when there are both a function and a function-like macro with the same name, and the programmer wants to prevent macro expansion.

This practice may seem a little odd at first, but the C library sets a precedent by providing some macros and functions with identical names.

One such function/macro pair is isdigit(). The library might define it as follows:

/* the macro */
#define isdigit(c) ...

/* the function */
int (isdigit)(int c) /* avoid the macro through the use of parentheses */
{
  return isdigit(c); /* use the macro */
}

Your function looks almost identical to the above, so I suspect this is what's going on in your code too.

The parantheses don't change the declaration - it's still just defining an ordinary function called foo.

The reason that they have been used is almost certainly because there is a function-like macro called foo defined:

#define foo(x) ...

Using (foo) in the function declaration prevents this macro from being expanded here. So what is likely happening is that a function foo() is being defined with its body being expanded from the function-like macro foo.

The parentheses are meaningless.
The code you show is nothing but an infinite recursion.

When defining a function pointer, you sometimes see strange parentheses that do mean something. But this isn't the case here.

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