函数重载和类型转换解析 [英] function overload and type conversion resolution

问题描述

为什么下面的代码没有显示对重载函数的未定义调用"错误?仅仅是因为int是内置类型?在标准中的哪里可以找到转换为内置类型的保证，例如在下面的代码中?...谢谢！

why do we not see a "undefined call to overloaded function" error with the code bellow? just because int is a built in type? where in the standard can I find the guarantee for the conversion to built in type, such as in the code bellow?... thanks!

#include <iostream>
using namespace std;

class B {
public:
operator int(){ return 0; }
};

class A {
public:
A( int i ) { };
};

void f ( int i ) { cout << "overload f(int) was used!";};
void f ( A a )   { cout << "overload f(A) was used!" ;};


int main () {
  B b;
  f( b );
}

推荐答案

与内置类型无关.您为B定义了operator int.这意味着您提供了从B到int的用户定义的转换.根据标准的12.3.4，最多将一个用户定义的转换(构造函数或转换函数)隐式应用于单个值".这就是为什么不将其转换为A的原因，因为这将需要两次隐式转换.

It has nothing to do with being a built-in type. You defined operator int for B. This means that you have provided a user-defined conversion from B to int. According to 12.3.4 of the standard, "at most one user-defined conversion (constructor or conversion function) is implicitly applied to a single value." This is why it is not converted to A, because that would require two implicit conversions.

确切确定何时发生这种情况的规则有些复杂，因此许多人建议您避免提供用户定义的转换.定义它们的另一种方法是为构造函数提供一个参数.您可以在开头添加explicit，以避免将其隐式应用.

The rules for determining exactly when this happens are somewhat complicated, so many people advise that you avoid providing user-defined conversions. Another way of defining these is to provide a constructor with one argument; you can add explicit to the beginning to avoid it being applied implicitly.

当您调用f(b)时，编译器将应用您提供的将b转换为int的转换.如果要将其转换为A，则必须定义从B到A的转换，或显式应用其中的一种转换，例如f(int(b))或f(A(b)).

When you call f(b), the compiler applies the conversion that you provided to convert b to int. If you want to convert it to A, you'll have to define a conversion from B to A, or apply one of the conversions explicitly, like f(int(b)) or f(A(b)).

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