In Java, when using bitshifts, why does 1 << 32 != 1 << 31 << 1?

Question

int a = 1 << 32;
int b = 1 << 31 << 1;

Why does a == 1? b is 0

Answer 1

There is some difference in how processors implement shift instructions.

For instance, IIRC, ARM processors (32-bit ISA) take the least significant byte of the shifting register. (Shifts are not actually standalone instructions on ARM).

So long as the underlying processor has a vaguely sensible way to shift, it's easier to clear all but the least significant bits (one instruction usually) than to check if the shift is large and branch (actually on the ARM this only typically adds one instruction because all instructions are conditional).