c++ - 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

但是在这种情况下，没有这样的转换。当我删除 explicit 时，代码可以正常工作。

然而，当我改变
显式运算符 int(){ return i; }到
显式运算符 bool(){ return i != 0; ，

并将if(s == 4)分别修改为if(s)，代码运行正常。

看起来转换规则有点困惑，有人可以更详细地解释一下吗？

Answer 1

其中implicit conversions , 自 C++11 起，将为 contextual conversions 考虑显式用户定义的转换函数;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.