How come pointer to a function be called without d

问题:

I have a weird typedef statement in a C++ program, generated by Py++.

double radius(int);  // function to be wrapped
typedef double (*radius_function_type)(int);    
bp::def("radius", radius_function_type(&radius));   // bp::def is a function for wrapping

What I figured out so far is that the above typedef statemnt is not of the type, most of us are familiar with,

typedef complex_type simple_alias;

Rather it is a way to declare pointer to a function which takes int as argument and returns double (same as the prototype). So my question now is that, how come pointer to a function (without dereferencing) be called with address of a function as an argument? This also doesn't match with the prototype. Somebody please explain!

回答1:

Your question is confusing. Are you asking what this does:

radius_function_type(&radius)"

This is just a C++ typecast, a bit like:

radius (int (42));

but since radius is already of type radius_function_type then you can just as easily do:

bp::def("radius", radius);

but as this is code generated by Py++, it's probably being extra careful with the output.

回答2:

It doesn't declare a function pointer variable but a function pointer typedef called radius_function_type. radius_function_type(&radius) is just a (redundant) cast for the function pointer itself. (The unary & address-of operator is also redundant; for a function, radius and &radius are the same thing.)

On a low level, calling a function is just placing the arguments somewhere according to the underlying calling convention (usually on the stack) and then jumping to a memory address. So the compiler can call a function with just a pointer if it knows the function pointer type (function signature) and the pointer value itself.

回答3:

Well ... It is a bit similar to how arrays are related to pointers, in C and C++. The name of a function is basically a pointer, too. Seen from that perspective, it's not too surprising that given a definition:

int foo(int a, int b)
{
  return a + b;
}

You can do the call either directly, through the pointer that is the function's name:

foo(1, 2);

or by storing that value in a separate variable, which must be declared as a pointer:

int (*pointer)(int, int) = foo;
pointer(1, 2);

In this case, we can call through the pointer variable directly, and there's also no need to explicitly "take the address of" the function by writing &foo.

回答4:

Dereferencing (in way you think) a function's pointer means: accessing a CODE memory as it would be a DATA memory.

Function pointer isn't suppose to be dereferenced in that way. Instead, it is called.

I would use a name "dereference" side by side with "call". It's OK.

Anyway: C is designed in such a way that both function name identifier as well as variable holding function's pointer mean the same: address to CODE memory. And it allows to jump to that memory by using call () syntax either on an identifier or variable.

回答5:

Function pointer is basically the address of the function that needs to be invoked. To call the function pointed to by a function pointer, you consider the function pointer as though it were the name of the function that you wish to call. The act of calling it itself performs the dereference; there is no need for explicit dereference.

Function pointer syntax can look like pointers (with & and *) or that can be omitted as well. As @unwind already pointed out it is similar to how arrays are treated where bare array is similar to pointer and optionally you can prefix the array with & to get the address.