C++ True numeric Pitfall, with Pow()

[VladLio]

#include
#include
#include
using namespace std;

int main() {
unsigned long long Bit1{ 0 };
Bit1 = (unsigned long long) (pow(2, 64) - 1);
println("(2^64)-1 = |{:64b}|\n", Bit1);
Bit1 = (unsigned long long) (pow(2, 63) - 1);
println("(2^63)-1 = |{:64b}|\n", Bit1);
Bit1 = (unsigned long long) (pow(2, 32) - 1);
println("(2^32)-1 = |{:64b}|\n", Bit1);
}
Result :
(2 ^ 64) - 1 = |10000000.00000000.00000000.00000000.00000000.00000000.00000000.00000000|
(2 ^ 63) - 1 = |10000000.00000000.00000000.00000000.00000000.00000000.00000000.00000000|
(2 ^ 32) - 1 = |11111111.11111111.11111111.11111111|

[Shriharsh-Deshmukh]

The issue arises from how pow() works and formatting the output as binary.

To resolve this:

1. Replace pow(2, n) with bitwise shifts for accurate integer representation.
2. Use C++ standard library bitset for binary formatting.

[VladLio]

ho ! Great, it's a clearer solution, that I did not know. I have made a partial shift with 2^32, and a sum, to solve the problem.
(1ULL << 64) there is a "surprising" no-overflow with this 64 shift, with " 1 ",
I might have thought 64 times means 1 on 65 the bit, perhaps back management ? But that's Ok !
Glad to have an echo in my publications ! Thanks .. there'll be over discuss and show.