decided to panic with beaten here. So: I wanted to withdraw all bits of the whole, but not dividing, but beaten.

string s = "";
  int x = 67542;
  For (int i = Sizeof (int) * 8-1; i & gt; = 0; i--)
    S + = ((X & amp; 1 & lt; & lt; i)? '1': '0');
  COUT & LT; & LT; s;

with int everything works very well. Displays 00000000000001000001111101010110
But something is wrong with Long Long (Signed / unsigned). Displays 0000000000000100000111110101111110101100000000000000000000011111010110
It seems like a number repeats twice. In theory, it should bring a bunch of zeros in front.

string s = "";
  Long Long unsigned x = 67542;
  For (int i = SizeOf (Long Unsigned) * 8-1; i & gt; = 0; i--)
    S + = ((X & amp; 1 & lt; & lt; i)? '1': '0');
  COUT & LT; & LT; s;

Main question: why is it going on?

Answer 1, Authority 100%

Here

(x & amp; 1 & lt; & lt; i)

The compiler believes that 1 you have int . And should be unsigned long long , otherwise it will simply be UB with large I .

(x & amp; 1ull & lt; & lt; i)

Try so …

Answer 2

Code:

std :: uint64_t number = 0xffffffff
For (int shift = 63; SHIFT & GT; = 0; --Shift)
  STD :: COUT & LT; & LT; ((Number & gt; & gt; shift) & amp; 1);

Output:

0000000000000000000000000000000011111111111111111111111111111111

Template:

template & lt; TypeName T & GT;
Concept Integral = STD :: is_integral_v & lt; T & GT ;;
Template & lt; Integral T & GT;
Void Binrep (Const T Number)
{
  for (int shift = sizeof (t) * 8 - 1; shift & gt; = 0; --shift)
    STD :: CLOG & LT; & LT; ((Number & gt; & gt; shift) & amp; 1);
  STD :: CLOG & LT; & LT; STD :: ENDL;
}