'<< Verilog中的'运算符 [英] ' << ' operator in verilog

问题描述

我有一个verilog代码，其中的一行如下:

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

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

此处将存储在RAM_DEPTH中的内容以及<<运算符在此处的作用.

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

推荐答案

<<是二进制移位，将1左移8位.

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

4'b0001 << 1 => 4'b0010

>>是将MSB加0的二进制右移.
>>>是有符号移位，如果左输入是有符号的，则它将保持MSB的值.

>> 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

三种指示左操作数的方式:

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

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