布尔lambda？ [英] Boolean lambdas?

问题描述

此代码如何编译???

How come this code compiles???

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;
}

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！

如果您引用此答案 ^[1] ，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^[1], non-capturing lambdas are convertible to function pointers. And it turns out again that function pointers, being pointers themselves, are implicitly convertible to bool!

4.12布尔转换[conv.bool]

¹ 算术，无范围枚举，指针或指向成员类型的指针可以转换为bool类型的
prvalue。零值，空指针值或空成员指针值将转换为false;
任何其他值都将转换为true。可以将类型std :: nullptr_t的prvalue转换为
类型的prval值bool;结果值为false。

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.

LIVE CODE

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

    bool b = lambda;

    if(lambda) {}
}

---

^[1] _{老实说，给了我写这个的想法。}

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