为什么调用了不合适的重载函数？

Question

为什么下面的代码总是打印"type is double"？（我在 Whosebug 上看到过这段代码）

#include <iostream>


void show_type(...) {
    std::cout << "type is not double\n";
}

void show_type(double value) { 
    std::cout << "type is double\n"; 
}

int main() { 
    int x = 10;
    double y = 10.3;

    show_type(x);
    show_type(10);
    show_type(10.3);
    show_type(y);


    return 0;
}

Answer 1

http://en.cppreference.com/w/cpp/language/overload_resolution 说：

A standard conversion sequence is always better than a user-defined conversion sequence or an ellipsis conversion sequence.

Answer 2

   void show_type(double value) { 
        std::cout << "type is double\n"; 
    }

如果您在行上方发表评论，则输出将为

type is not double
type is not double
type is not double
type is not double

这意味着编译总是优先选择 void show_type(double value) 而不是 void show_type(...)。

在你的情况下，如果你想调用方法 void show_type(...) 在调用此方法时传递两个或更多参数 show_type(firstParameter,secondParameter)

#include <iostream>


void show_type(...) {
    std::cout << "type is not double\n";
}

void show_type(double value) { 
    std::cout << "type is double\n"; 
}

int main() { 
    int x = 10;
    double y = 10.3;

    show_type(x);
    show_type(10);
    show_type(10.3);
    show_type(y);
    show_type(4.0,5); //will called this  method show-type(...) 


    return 0;
}

现在上面一行的输出将是

type is  double
type is  double
type is  double
type is  double
type is not double  //notice the output here

more info on var-args

为什么调用了不合适的重载函数？

Why is the unsuitable overloaded function is called?

c++

overloading