c陷阱与缺陷第二章——Syntatic Pitfalls

2.1 Understanding function declarations

Every C variable declaration has two parts: a type and a list of expression-like things called declarators.

A declarator looks something like an expression that is expected to evaluate to the given type.

For instance:

float f, g;

indicates that the expressions f and g, when evaluated, will be of type float.

Because a declarator looks like an expression, parentheses may be used freely:

float ((f));

means that ((f)) evaluates to a float and therefore, by inference, that f is also a float.

Analogously,

float *pf;

means that *pf is a float and therefore that pf is pointer to a float.

Once we know how to declare a variable of a given type, it is easy to write a cast for that type: just remove the variable name and the semicolon from the declaration and enclose the whole thing in parentheses.

Thus, since

float (*h)();

declares h to be a pointer to a function returning a float,

(float(*)())

is a cast to a pointer to a function returning a float.

Now we can analyze the following statement:

(*(void(*)())0) ();

First, we can see that

(void(*)())

is a cast to a pointer to function returning void.

So, $0$ is cast to such a pointer.

Thus, we can analyze the statement as follows:

(*fp)();

fp is a pointer to a function returning void, *fp is the function itself, so (*fp)() is the way to invoke it.