Question

When constructor is explicit, it isn’t used for implicit conversions. In the given snippet, constructor is marked as explicit. Then why in case foo obj1(10.25); it is working and in foo obj2=10.25; it isn’t working ?

#include <iostream>
class foo
{
    int x;
    public:
    explicit foo( int x ):x(x)
    {}
};

int main()
{
    foo obj(10.25);  // Not an error. Why ?
    foo obj2 = 10.25; // Error
    getchar();
    return 0;
}

error: error C2440: ‘initializing’ : cannot convert from ‘double’ to ‘foo’

Answer 1

There is a difference between these two lines of code. The first line,

foo obj(10.25);

explicitly calls your foo constructor passing in 10.25. This syntax, in general, is a constructor call.

The second line,

foo obj2 = 10.25;

tries to implicitly convert from 10.25 to an object of type foo, which would require use of an implicit conversion constructor. In this example, you’ve marked the constructor explicit, there’s no implicit conversion constructor available, and so the compiler generates an error.