conversion operator as standalone function

c++ type-conversion conversion-operator
shura picture shura · Jan 31, 2010 · Viewed 7.8k times · Source

Why does C++ require that user-defined conversion operator can only be non-static member? Why is it not allowed to use standalone functions as for other unary operators? Something like this:

operator bool (const std::string& s) { return !s.empty(); }

Answer

anon picture anon · Jan 31, 2010

The one reason I can think of is to prevent implicit conversions being applied to the thing being cast. In your example, if you said:

 bool( "foo" );

then "foo" would be implicitly converted to a string, which would then have the explicit bool conversion you provided applied to it.

This is not possible if the bool operator is a member function, as implicit conversions are not applied to *this. This greatly reduces the possibilities for ambiguity - ambiguities normally being seen as a "bad thing".

Related questions

Conversion constructor vs. conversion operator: precedence

Reading some questions here on SO about conversion operators and constructors got me thinking about the interaction between them, namely when there is an 'ambiguous' call. Consider the following code: class A; class B { public: B(){} B(const A&) //…

Read More
Easiest way to convert int to string in C++

What is the easiest way to convert from int to equivalent string in C++. I am aware of two methods. Is there any easier way? (1) int a = 10; char *intStr = itoa(a); string str = string(intStr); (2) int a = 10; stringstream ss; ss &…

Read More
How to convert string to char array in C++?

I would like to convert string to char array but not char*. I know how to convert string to char* (by using malloc or the way I posted it in my code) - but that's not what I want. I …

Read More