How to define strong ID types in C++11?

Question

How to define strong ID types in C++11? It's posible to done alias of integer types but getting warnings from compiler when you mix types?

E.g:

using monsterID = int;
using weaponID = int;

auto dragon = monsterID{1};
auto sword = weaponID{1};

dragon = sword; // I want a compiler warning here!!

if( dragon == sword ){ // also I want a compiler warning here!!
    // you should not mix weapons with monsters!!!
}

You can't do this with typedefs. Make new types. There's something in Boost IIRC. — R. Martinho Fernandes, Aug 20 '13 at 14:54
Of interest is this proposed new feature for C++1y: http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2013/n3515.pdf — Andrew Tomazos, Aug 20 '13 at 15:08
@user1131467, so it's not posible to do it now, without using a own class. — Zhen, Aug 20 '13 at 16:23

score 5 · Accepted Answer · answered Aug 20 '13 at 14:56

5

If youre using boost, try BOOST_STRONG_TYPEDEF

Example from the documentation:

BOOST_STRONG_TYPEDEF(int, a)
void f(int x);  // (1) function to handle simple integers
void f(a x);    // (2) special function to handle integers of type a 
int main(){
    int x = 1;
    a y;
    y = x;      // other operations permitted as a is converted as necessary
    f(x);       // chooses (1)
    f(y);       // chooses (2)
}

answered Aug 20 '13 at 14:56

Viktor Sehr

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Thanks. I think it's a great solution. But I was looking if with new standard C++11 it's posible to do it without using macros, or fully classes. – Zhen Aug 20 '13 at 16:14

How to define strong ID types in C++11?

1 Answers1