Question

Here is the code:

#include <iostream>

using namespace std;

template<class T>
void printbinary3(T num){
   for(int i = sizeof(T) * 8 - 1; i >= 0 ; --i) {
      if((1 << i) &  num)
         cout << "1";
      else
         cout << "0";   
   }
   cout << endl;
}

int main ()
{
   char a = 1;
   short b = 1;
   int c = 1;
   long d = 1;
   __int64 e = 1;
   unsigned __int64 f = 1;

   printbinary3(a);
   printbinary3(b);
   printbinary3(c);
   printbinary3(d);
   printbinary3(e);
   printbinary3(f);
   return 0;
}   

Here is the output:

00000001
0000000000000001
00000000000000000000000000000001
00000000000000000000000000000001
0000000000000000000000000000000100000000000000000000000000000001

When compiling I get this error for the unsigned __int64 variable – f.

1> : warning C4334: '<<' : result of 32-bit shift implicitly converted to 
     64 bits (was 64-bit shift intended?)
1>   xx.cpp(66) : see reference to function template instantiation 
     'void printbinary3<unsigned __int64>(T)' being compiled

Why am I not seeing

0000000000000000000000000000000000000000000000000000000000000001

as the output for the 64 bit integer?

Angus

Answer 1

In your printbinary3 function:

void printbinary3(T num){
   for(int i = sizeof(T) * 8 - 1; i >= 0 ; --i) {
      if((1 << i) &  num)

Here, you are creating a literal 1 of type int, which most probably is 32 bit. If T = __int64, you will shift it for too many bits, leading to an overflow and the warning.

Instead of using implicit int, explicitely create a T instead:

      if((T(1) << i) &  num)

This should fix the warning and the output.