Typedef function pointer in C++

Question

I am trying to understand a code in c++ header.

#define GET_VAL(fn) void fn(int val)
typedef GET_VAL ((*get_val));
struct myStruct
{ 
    get_val getValue;
};

In the source file, the function getValue is called.

getValue(2);

Anyone have any ideas?

This is a horrible way to declare a function pointer. I don't see a reason to use the macro like that other than obfuscation. — interjay
– interjay, Commented Mar 11, 2015 at 15:54
It is a piece of code in a big project. I just gave a simplified version. Maybe they have their own consideration. Anyway you are right, it did obfuscate me. — John
– John, Commented Mar 12, 2015 at 13:18

TartanLlama · Accepted Answer · 2015-03-11 15:54:31Z

8

The GET_VAL macro substitutes the tokens you pass to it. This:

typedef GET_VAL ((*get_val));

Expands to:

typedef void (*get_val) (int val);

That is a pointer to a function which takes an int and returns nothing. A function pointer of this type is declared in myStruct (which is presumably defined at some point) and called like a regular function.

answered Mar 11, 2015 at 15:54

TartanLlama

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1 Comment

John Over a year ago

Thanks Tartan. Solved my confusion.

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