CODEWITHSUNDEEP
c++
iPherian
iPherian

Reputation: 958

Overloaded template function doesn't select the correct version for a specific type

Situation is: I have a generic function but would like an altered version for a specific type.

I would instinctively write a widget const& w parameter to be able to accept any widget by reference, but that doesn't compile in that case, because the compiler ignored the specific overload (2) and used generic (1) instead. It does compile if I remove the const in the params of (2), why is that?

godbolt

struct widget {
    int widget_size;
};

// (1)
// compiler uses this instead of (2)
template <typename Arg>
int get_size(Arg && arg) {
   return arg.size;
}

// (2)
int get_size(widget const& arg) {
    return arg.widget_size;
}

int main() {
    widget w;
    get_size(w);
}

Upvotes: 0

Views: 45

Answers (1)

H S
H S

Reputation: 1221

because the compiler ignored the specific overload .

Please note that Version 2 is a better match as argument passed lacks const keyword and version 1 also lacks const keyword.

that doesn't compile in that case.

Widget class doesn't have size member variable in it. It's got widget_size , which is causing compilation error. 

int get_size(Arg && arg) {
   return arg.widget_size; // See this
}

It does compile if I remove the const in the params of (2), why is that?

This is because get_size(w); started matching version 2, thereby omitting any need for compiler to check for widget_size in version 1.

Upvotes: 4

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