CODEWITHSUNDEEP
c++operator-overloading
Ryan Brown
Ryan Brown

Reputation: 1095

C++ 2 overloads have similar conversions

I'm writing wrapper classes for basic types so they can be typed... except I've run into a problem. I need both these methods to make it nice:

Integer32(FastInteger32 value);
Boolean operator>(Integer32 value);
operator FastInteger32();

Except when I go to use the '>' operator between a 'FastInteger' and an 'Integer', the compiler sees two possible routes to go and doesn't just pick one, it blows up. Is there a way to to tell it to just pick one?

The entire code:

typedef unsigned int FastInteger32;

class Integer32
{
public:
    Integer32(FastInteger32 value);
    Boolean operator>(Integer32 value);
    operator FastInteger32();
private:
    FastInteger32 value;
};

int main()
{
    Integer32 a = 5;
    FastInteger32 b = 5;
    if (a < b) { }  // Doesn't know what to do here, convert b to Integer32 and compare or convert a to FastInteger32 and compare
}

You guys rock by the way, thanks for helping!

Upvotes: 1

Views: 2142

Answers (2)

MSalters
MSalters

Reputation: 180135

C++ will always prefer an overload which doesn't have a conversion at all. Therefore, adding bool operator<(FastInteger32, Integer32) and bool operator<(Integer32, FastInteger32) will eliminate all ambiguities.

Upvotes: 1

Alex Chamberlain
Alex Chamberlain

Reputation: 4207

The problem is with your constructor (Integer32(FastInteger32 value);). Single argument constructors should (nearly) always be declared as explicit. Simply change the declaration to explicit Integer32(FastInteger32 value);; the definition should not change.

Upvotes: 1

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