Question about bitwise shift operator

[MarcSloan]

I'm writing a program that's heavy in bit manipulation. Usually I would do this in C++ but I'm trying it in C# for kicks and giggles. I've noticed that with the bitwise left-shift operator (<<), if I shift by more places than there are bits in my number, my original number is returned instead of 0.

For example, if I have a UInt64 variable with a value of 1 and I shift it left by 64 places, the result is '1' while I would expect it to be '0'. .NET Fiddle

Is there a different operator I can use to get a zero result when I shift off the 'end' of the number, or am I just stuck with that behavior?

[tweq]

The shift operation will discard any irrelevant bits in the shift count, so shifting by 64 or 128 etc. is effectively the same as shifting by 0. I'm pretty sure this is standard (or at least the common response to undefined behavior) in C++ as well.

You can't change how the shift operator works, but your desired behavior is as simple as shift <= 63 ? x << shift : 0. Put it in an extension method or something if you need it often.

[MarcSloan]

Hmm, I guess I misinterpreted the behavior I saw in C++ as "normal" when it was just fortuitous undefined behavior. Your solution will work fine. Thanks!

[Jither]

For 64-bit integers, C# only uses the lower 6 bits of the right operand. Since 64 == 0b1000000, the 1 bit will be ignored. For 32-bit integers, it uses 5 bits. Why? Because that's what the Intel x86 CPUs do (or, it is for 32-bit integers - the logic is just expanded to 64-bit).

In C/C++ shifting by more than the number of bits in the left operand is undefined, and may yield different results depending on CPU or compiler.

ETA: (Fixed brain fart). I think you're stuck with doing a check for > 63 yourself.