c++ – Typedef function pointer?

c++ – Typedef function pointer?

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.


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
  1. typedef is used to alias types; in this case youre aliasing FunctionFunc to void(*)().

  2. Indeed the syntax does look odd, have a look at this:

    typedef   void      (*FunctionFunc)  ( );
    //         ^                ^         ^
    //     return type      type name  arguments
  3. No, this simply tells the compiler that the FunctionFunc type will be a function pointer, it doesnt define one, like this:

    FunctionFunc x;
    void doSomething() { printf(Hello theren); }
    x = &doSomething;
    x(); //prints Hello there

c++ – Typedef function pointer?

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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