Question

When it comes to constructors, adding the keyword explicit prevents an enthusiastic compiler from creating an object when it was not the programmer’s first intention. Is such mechanism available for casting operators too?

struct Foo
{
    operator std::string() const;
};

Here, for instance, I would like to be able to cast Foo into a std::string, but I don’t want such cast to happen implicitly.

Answer 1

Yes and No.

It depends on which version of C++, you’re using.

• C++98 and C++03 do not support explicit type conversion operators
• But C++11 does.

Example,

struct A
{
    //implicit conversion to int
    operator int() { return 100; }

    //explicit conversion to std::string
    explicit operator std::string() { return "explicit"; } 
};

int main() 
{
   A a;
   int i = a;  //ok - implicit conversion 
   std::string s = a; //error - requires explicit conversion 
}

Compile it with g++ -std=c++0x, you will get this error:

prog.cpp:13:20: error: conversion from ‘A’ to non-scalar type ‘std::string’ requested

Online demo : http://ideone.com/DJut1

But as soon as you write:

std::string s = static_cast<std::string>(a); //ok - explicit conversion 

The error goes away : http://ideone.com/LhuFd

BTW, in C++11, the explicit conversion operator is referred to as "contextual conversion operator" if it converts to boolean. Also, if you want to know more about implicit and explicit conversions, read this topic:

• Implicit VS Explicit Conversion