为什么会出现从指针隐式类型转换在C ++中为bool？ [英] Why is there an implicit type conversion from pointers to bool in C++?

问题描述

考虑级富像这样定义了两个构造函数：

Consider the class foo with two constructors defined like this:

class foo
{
public:
    foo(const std::string& filename) {std::cout << "ctor 1" << std::endl;}
    foo(const bool some_flag = false) {std::cout << "ctor 2" << std::endl;}
};

实例化一个字符串类，并猜测其构造函数被调用？

Instantiate the class with a string literal, and guess which constructor is called?

foo a ("/path/to/file");

输出：

男星2

我不知道你怎么想，但我不发现，在节目历史上最直观的表现。我敢打赌，有一些聪明的原因吧，不过，我想知道是什么可能？

I don't know about you, but I don't find that the most intuitive behavior in programming history. I bet there is some clever reason for it, though, and I'd like to know what that might be?

推荐答案

这是很常见的用C写这个

It's very common in C to write this

void f(T* ptr) {
    if (ptr) {
        // ptr is not NULL
    }
}

您应该做一个为const char * 构造。

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