Arithmetic shift acts as a logical shift, regardless of the signed variable

问题

I've got a register declared as so:

logic signed [15:0][2:0][15:0] registers;

When I place a 2's compliment number into the array and arithmetically shift the number, it logical shifts instead:

registers[0][0] = 16'b1000000000000000;
registers[0][0] = registers[0][0]>>>2;

Apparently, the system will logical shift instead of arithmetically shift if the number is not signed. However as you can clearly see, 'registers' is definitely signed.

Does anybody know what I might be missing here?

Thanks!

回答1:

With Verilog, once you take a part-select, the result is unsigned. Use the $signed system task on the part select to make it signed.

res = $signed(registers[0][0]) >>> 2;

来源：https://stackoverflow.com/questions/14197117/arithmetic-shift-acts-as-a-logical-shift-regardless-of-the-signed-variable