为什么float参数适合int函数参数？ [英] Why does float argument fit to int function parameter?

问题描述

请查看以下代码：

#include <iostream>
class A {
  public:
    int my;
    A(int a=0) : my(a) { }
};

int main() {
  A x = 7; // 1
  A y = 6.7; // 2
  std::cout << x.my << " " << y.my << "\n";
}

尽管没有 A（double a）; 构造函数。
当完全允许编译器将一种参数类型转换为另一种参数类型以调用相应的构造函数时？

It actually compiles although there is no A(double a); constructor. When exactly compiler is allowed to convert one argument type to another to call corresponding constructor?

推荐答案

cppreference具有标准转化列表。您感兴趣的是浮动-积分转换部分，该部分也可以在 N4140 4.9 / 1

cppreference has a list of standard conversions. Of interest to you is the Floating - integral conversions section which can also be found in N4140 4.9/1

浮点类型的prvalue可以转换为任何整数类型。小数部分被截断，即小数部分被丢弃。

A prvalue of floating-point type can be converted to prvalue of any integer type. The fractional part is truncated, that is, the fractional part is discarded.

查找 A（int）要通过标准转换进行调用，编译器会插入必要的步骤以使代码正常工作。相同的规则允许 int x = 1.1 进行编译

Finding A(int) to be callable with a standard conversion, the compiler inserts the necessary step to make the code work. It's the same rule that allows int x = 1.1 to compile

如果这种行为不受欢迎，您可以使用 =删除

If this behavior is undesirable you can forbid it with an =delete

class A {
  public:
    //...
    A(int a);
    A(double) =delete;
};    

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