On Tuesday, 18 December 2018 at 20:33:43 UTC, Rainer Schuetze
wrote:
On 14/12/2018 02:56, Steven Schveighoffer wrote:
On 12/13/18 7:16 PM, Michelle Long wrote:
byte x = 0xF;
ulong y = x >> 60;
Surely you meant x << 60? As x >> 60 is going to be 0, even
with a ulong.
It doesn't work as intuitive as you'd expect:
void main()
{
int x = 256;
int y = 36;
int z = x >> y;
writeln(z);
}
prints "16" without optimizations and "0" with optimizations.
This happens for x86 architecture because the processor just
uses the lower bits of the shift count. It is probably the
reason why the language disallows shifting by more bits than
the size of the operand.
Yes. On x86 shifting (x >> y) is in reality x >> (y & 0x1F) on 32
bits and x >> (y & 0x3F) on 64 bits.
