为什么在 C++ 中存在从指针到 bool 的隐式类型转换? [英] Why is there an implicit type conversion from pointers to bool in C++?

问题描述

考虑类 foo 有两个这样定义的构造函数:

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;}
};

用字符串字面量实例化类，猜猜调用的是哪个构造函数?

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

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

输出:

构造函数 2

我不了解您，但我不认为这是编程史上最直观的行为.不过，我敢打赌它有一些巧妙的理由，我想知道那可能是什么?

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

在 C 中写这个很常见

It's very common in C to write this

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

你应该创建一个 const char* 构造函数.

You should make a const char* constructor.

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