c++ - bool lambda？

Question

这段代码怎么编译？？？

LIVE CODE

#include <iostream>


int main() {
    auto lambda1 = []{};
    auto lambda2 = []{};

    if(lambda1 && lambda2) {
        std::cout << "BOOLEAN LAMBDAS!!!" << std::endl;
    }

    if(lambda1 || lambda2) {
        std::cout << "BOOLEAN LAMBDAS AGAIN FTW!!!" << std::endl;
    }

    bool b1 = lambda1;
    bool b2 = lambda2;

    std::cout << b1 << ", " << b2 << std::endl;
}

bool lambda! (或者 boolambdas，如果你愿意... *回避*)

这怎么行？这是另一个 GCC 错误吗？如果不是，这是标准吗？

Answer 1

原来是标准的!

如果你引用this answer ^[1]，非捕获 lambda 可转换为函数指针。事实证明，函数指针本身就是指针，可以隐式转换为 bool!

4.12 Boolean conversions [conv.bool]

¹ A prvalue of arithmetic, unscoped enumeration, pointer, or pointer to member type can be converted to a prvalue of type bool. A zero value, null pointer value, or null member pointer value is converted to false; any other value is converted to true. A prvalue of type std::nullptr_t can be converted to a prvalue of type bool; the resulting value is false.

为了证明转换为函数指针是所有这一切发生的原因，我尝试用捕获 lambda 来做同样的事情。然后生成“无法转换为 bool”错误。

LIVE CODE

int main() {
    int i;
    auto lambda = [i]{};

    bool b = lambda;

    if(lambda) {}
}

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^[1] _{老实说，这给了我写这篇文章的想法。}