参考文档

user-defined conversion function - cppreference.com
The Safe Bool Idiom - 知乎

一般形式为operator *type*() const, 比如:
operator int() const;
operator bool() const;
operator AA() const;

自定义类型转换

struct To
{
    To() = default;
    To(const struct From&) {} // converting constructor
};

struct From
{
    operator To() const {return To();} // conversion function
};

int main()
{
    From f;
    To t1(f);  // direct-initialization: calls the constructor
    // Note: if converting constructor is not available, implicit copy constructor
    // will be selected, and conversion function will be called to prepare its argument

//  To t2 = f; // copy-initialization: ambiguous
    // Note: if conversion function is from a non-const type, e.g.
    // From::operator To();, it will be selected instead of the ctor in this case

    To t3 = static_cast<To>(f); // direct-initialization: calls the constructor
    const To& r = f;            // reference-initialization: ambiguous
}

为什么operator bool()需要用explicit修饰?

c++ - Why does declaring an operator bool() const member overload the [] operator? - Stack Overflow

The operator is coming from the built-in subscript operator which treats expressions A[B] as *(A + B).
This results in the evaluation of *(1 + "wut") => 'u', which then causes the if condition to pass, as 'u' is a non-zero value.
Declare your member as explicit operator bool() to prevent your type from being implicitly converted to other integral types.

#include <iostream>
using namespace std;

struct Test {
    operator bool() const {
        return true;
    }
};

int main(int argc, char** argv) {
    Test test;

    if (test["wut"])
        cout << "Success (test[\"wut\"])\n";
}

一个operator bool()的坑

c++ - Why is my "explicit operator bool()" not called? - Stack Overflow