Why the overloaded function template for the float parameter does not call the overloaded ordinary float function?

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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;
         }

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

Posted 17-Aug-11 0:10am

Ashutosh_g

Updated 17-Aug-11 0:26am

Prerak Patel

v4

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Orjan Westin · Answer 1 · 2011-08-17T00:28:00

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>

is called.

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

display( float('c') )

function is called.

వేంకటనారాయణ(venkatmakam) · Answer 2 · 2011-08-17T00:30:00

Solution 2

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.

Posted 17-Aug-11 0:30am

వేంకటనారాయణ(venkatmakam)

Comments

Philippe Mori 17-Aug-11 23:41pm

You should do an overload... Otherwise, everytime, the cast if forgotten, the wrong function would be called. And, by the way, we should avoid cast whenever possible.