> However, consider this:
>
> input [31:0] x;
> input [5:0] shift;
> output [31:0] z;
> wire [63:0] y = {x, 31'b0};
> assign z = y >> (shift + 32);
>
> The shift+32 is cheap, because all you're really doing is inverting
> the high bit of the shift amount. The only question is how much more
> expensive is a 64-bit shifter than a 32-bit shifter.
How does this compare to:
z = shift[5] ? (y >> shift[4:0]) : (y << shift[4:0])
If you turned this into a switch statement you could also use the high bits of
the shift count to select between arithmetic and logical shifts.
For the compiler size of things, or-ing a few high bits is as easy as negating
the shift count.
Paul
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