为什么float参数的重载函数模板没有调用重载的普通float函数? [英] Why the overloaded function template for the float parameter does not call the overloaded ordinary float function?

问题描述

这是我的代码:

Here is my code:

#include<iostream.h>
#include<conio.h>

template <class t="">
 void display(T x)
         {
              cout<<"Template display :"<<x<<"\n";
         }
         
         
          void display(float x)                                  
         {
              cout<<"Explicit display: float "<<x<<"\n";
         }
       
         
         
         int main()
         {
            
             
             display(12.34);
             display(''c'');
             getch();
             return 0;
         }

对于display(12.34)，调用了模板函数，但是在int的情况下，调用了正常的重载显示函数，为什么呢?

For display(12.34), the template function is called but in case of int the normal overloaded display function is called, why?

推荐答案

12.34是double而不是a浮点数，并且双精度数不能以确保无损失的方式转换为浮点数.因此，将调用模板功能

12.34 is a double, not a float, and doubles can''t be converted to floats in a way that is guaranteed to be loss-free. Therefore, the template function

display<double>(12.34)</double>

.

"c"是一个字符，可以自动转换为浮点数而不会丢失，因此调用了

is called.

''c'' is a char, which can be automatically converted to a float without loss, and therefore the

display( float(''c'') )

函数.

通常将值12.34视为double value.这就是为什么不调用显式函数的原因.因此，您可以将12.34强制转换为float或将显式函数用于double.

Normally value 12.34 will be considered as double value.That is why your explicit function is not called.so you can cast the 12.34 to float or use explicit function for double.

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