编译器无法将显式转换应用于用作条件的表达式

Question

这是我的代码：

#include <iostream>

using namespace std;
class Sales{
  public:
    Sales(int i = 0):i(i){}
    explicit operator int(){ return i; }
  private:
    int i;
};

int main(){
  Sales s(5);
  if(s == 4) cout << "s == 4" << endl;
  else cout << "s != 4" << endl;
  return 0;
}

在c++ primer(5th)中，它说：

compiler will apply an explicit conversion to an expression used as a condition

但在这种情况下，没有这种转换。当我删除 explicit 时，代码可以正常工作。

然而，当我改变
explicit operator int(){ return i; } 至
explicit operator bool(){ return i != 0; }、

并分别将 if(s == 4) 更改为 if(s)，然后代码可以正常工作。

看起来转换规则有点混乱，有人可以更详细地解释一下吗？

Answer 1

在implicit conversions, since C++11, the explicit user-defined conversion function will be considered for contextual conversions之中；这发生在以下上下文中，并且需要类型 bool 时。对于 int，它不适用。

In the following five contexts, the type bool is expected and the implicit conversion sequence is built if the declaration bool t(e); is well-formed. that is, the explicit user-defined conversion function such as explicit T::operator bool() const; is considered. Such expression e is said to be contextually convertible to bool.
controlling expression of if, while, for;
the logical operators !, && and ||;
the conditional operator ?:;
static_assert;
noexcept.

编译器无法将显式转换应用于用作条件的表达式

compiler failed to apply explicit conversion to an expression used as a condition

c++

implicit-conversion

explicit-conversion

c++11