CODEWITHSUNDEEP
c++c++11
user3853544
user3853544

Reputation: 583

C++ function pointer syntax. Why does (*) work but * not?

I am writing a program for an assignment for a post fix calculator. There are bonus points if I can do it without using an if statement (no switch or while or any comparisons). I was reading about function pointers and came up with the following code:

int main(){
    unordered_map<char,int*(int,int)> m;
    m.insert({'+',[](int const a, int const b){return a+b;}});
}

This doesn't work.

However this does:

int main(){
    unordered_map<char,int(*)(int,int)> m;
    m.insert({'+',[](int const a, int const b){return a+b;}});
}

Why are (*) and * different?

Upvotes: 2

Views: 177

Answers (2)

vsoftco
vsoftco

Reputation: 56557

The type of 

int* (int,int)

is function taking (int, int) and returning int*, i.e. a pointer to int. On the other hand, the type of

int (*) (int, int)

is pointer to function taking (int, int) and returning int. So they are different declarations. In your map you insert types that are decay-able to the latter, not the former, and that's why the latter code works but not the former.

Useful: http://www.cdecl.org/

Super useful: The clockwise/Spiral rule

Upvotes: 5

Jerry Coffin
Jerry Coffin

Reputation: 490148

Without the parens, the int* is parsed as just "pointer to int", and the rest of the code just doesn't work.

With the parens, int (*) means "pointer to function returning an int" instead, and the parens (and stuff they enclose) specify the arguments to that function, so it all parses as what you want.

Upvotes: 3

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