Typedef function pointer?
I'm learning how to dynamically load DLL's but what I don't understand is this line
typedef void (*FunctionFunc)();
I have a few questions. If someone is able answer them I would be grateful.
typedef used? void should there not be a function name or something? It looks like an anonymous function. So I'm confused at the moment; can you clarify things for me?
typedef is a language construct that associates a name to a type.
You use it the same way you would use the original type, for instance
typedef int myinteger;
typedef char *mystring;
typedef void (*myfunc)();
using them like
myinteger i; // is equivalent to int i;
mystring s; // is the same as char *s;
myfunc f; // compile equally as void (*f)();
As you can see, you could just replace the typedefed name with its definition given above.
The difficulty lies in the pointer to functions syntax and readability in C and C++, and the typedef can improve the readability of such declarations. However, the syntax is appropriate, since functions - unlike other simpler types - may have a return value and parameters, thus the sometimes lengthy and complex declaration of a pointer to function.
The readability may start to be really tricky with pointers to functions arrays, and some other even more indirect flavors.
To answer your three questions
Why is typedef used? To ease the reading of the code - especially for pointers to functions, or structure names.
The syntax looks odd (in the pointer to function declaration) That syntax is not obvious to read, at least when beginning. Using a typedef declaration instead eases the reading
Is a function pointer created to store the memory address of a function? Yes, a function pointer stores the address of a function. This has nothing to do with the typedef construct which only ease the writing/reading of a program ; the compiler just expands the typedef definition before compiling the actual code.
Example:
typedef int (*t_somefunc)(int,int);
int product(int u, int v) {
return u*v;
}
t_somefunc afunc = &product;
...
int x2 = (*afunc)(123, 456); // call product() to calculate 123*456
typedef is used to alias types; in this case you're aliasing FunctionFunc to void(*)() .
Indeed the syntax does look odd, have a look at this:
typedef void (*FunctionFunc) ( );
// ^ ^ ^
// return type type name arguments
No, this simply declares that the FunctionFunc type will be a function pointer, it doesn't define one, like this:
FunctionFunc x;
void doSomething() { printf("Hello theren"); }
x = &doSomething;
x(); //prints "Hello there"
Without the typedef word, in C++ the declaration would declare a variable FunctionFunc of type pointer to function of no arguments, returning void .
With the typedef it instead defines FunctionFunc as a name for that type.
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