'if (x)' 'x' 是 class 的实例，而不是隐式转换？

Question

根据cppreference.com，显式转换函数不能用于隐式转换。例如，他们有这个：

struct B
{
    explicit B(int) { }
    explicit B(int, int) { }
    explicit operator bool() const { return true; }
};
 
int main()
{
    ...
    if (b2) ;      // OK: B::operator bool()
    ...
}

本来以为'if (b2)'是隐式转换，所以不能使用显式转换函数。那么什么是不允许的隐式转换的示例？

Answer 1

Contextual conversions

In the following contexts, the type bool is expected and the implicit conversion is performed if the declaration bool t(e); is well-formed (that is, an explicit conversion function such as explicit T::operator bool() const; is considered). Such expression e is said to be contextually converted to bool.

the controlling expression of if, while, for;

...

Answer 2

来自 C++ 17 标准（7 个标准转换）

4 Certain language constructs require that an expression be converted to a Boolean value. An expression e appearing in such a context is said to be contextually converted to bool and is well-formed if and only if the declaration bool t(e); is well-formed, for some invented temporary variable t (11.6).

声明

bool t( b2 );

格式正确，因为使用了直接初始化。

'if (x)' 'x' 是 class 的实例，而不是隐式转换？

How is 'if (x)' where 'x' is an instance of a class, not an implicit conversion?

c++

if-statement

operator-overloading

explicit-conversion