布尔lambda？

问题描述

此代码如何编译???

How come this code compiles???

#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;
}

Boolean lambdas！ （或boolambdas，如果你会... *害羞*）

Boolean lambdas! (Or boolambdas, if you will... *shies away*)

这是如何工作的？这是另一个GCC错误吗？如果没有，是这个标准吗？

How come this works? Is this another GCC bug? If not, is this standard?

推荐答案

原来是标准的！

如果您引用 ，非 - 捕获 lambdas可转换为函数指针。再次证明，作为指针本身的函数指针可隐式转换为 bool ！

If you refer to this answer, non-capturing lambdas are convertible to function pointers. And it turns out again that function pointers, being pointers themselves, are implicitly convertible to bool!

为了证明转换为函数指针是什么使得所有这一切发生，我试图做同样的事情捕获lambdas。然后无法转换为 bool 错误。

To give a supporting proof that the conversion to function pointer is what makes all of this happen, I've tried doing the same thing with capturing lambdas. Then "can't convert to bool" errors are generated.

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

    bool b = lambda;

    if(lambda) {}
}

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