' << ' operator in verilog

问题:

i have a verilog code in which there is a line as follows:

parameter ADDR_WIDTH = 8 ;
parameter RAM_DEPTH = 1 << ADDR_WIDTH;

here what will be stored in RAM_DEPTH and what does the << operator do here.

回答1:

<< is a binary shift, shifting 1 to the left 8 places.

4'b0001 << 1 => 4'b0010

>> is a binary right shift adding 0's to the MSB.
>>> is a signed shift which maintains the value of the MSB if the left input is signed.

4'sb1011 >>  1 => 0101
4'sb1011 >>> 1 => 1101

Three ways to indicate left operand is signed:

module shift;
  logic        [3:0] test1 = 4'b1000;
  logic signed [3:0] test2 = 4'b1000;

  initial begin
    $display("%b", $signed(test1) >>> 1 ); //Explicitly set as signed
    $display("%b", test2          >>> 1 ); //Declared as signed type
    $display("%b", 4'sb1000       >>> 1 ); //Signed constant
    $finish;
  end
endmodule

回答2:

1 << ADDR_WIDTH means 1 will be shifted 8 bits to the left and will be assigned as the value for RAM_DEPTH.

In addition, 1 << ADDR_WIDTH also means 2^ADDR_WIDTH.

Given ADDR_WIDTH = 8, then 2^8 = 256 and that will be the value for RAM_DEPTH

回答3:

<< is the left-shift operator, as it is in many other languages.

Here RAM_DEPTH will be 1 left-shifted by 8 bits, which is equivalent to 2^8, or 256.