Reputation: 711

verilog bit shift with 1

I'm trying to bit shift a value in verilog such that the replaced bits are 1's instead of 0's. i.e. I want to do 0001 << 1 such that it gives 0011 instead of 0010

Upvotes: 3

Answers (7)

SysTom

Reputation: 21

x |= x<<1; // assumes x starts as 1

Upvotes: 0

Roopa Vikas

Reputation: 1

function bit[31:0] cal_mask_value (int size);
    bit [31:0] mask_value;

    mask_value = 32'hFFFFFFFF << size;
    mask_value = ~mask_value;
    return mask_value;
endfunction

Upvotes: 0

Poorvaja Sankaran

Reputation: 1

We can use this,

a = (1'b1 << count) - 1;

Assign count to a value equal to the no.of 1's to be filled from LSB.

Upvotes: 0

Matthew

Reputation: 13937

You could do this:

module ones_shift #(log2_width=2) (input [(2**log2_width)-1:0] A,  input [log2_width:0] SHIFT, output [(2**log2_width)-1:0] As);

  wire [(2**log2_width)-1:0] Ai, Ais;

  assign Ai = ~A;
  assign Ais = Ai << SHIFT;
  assign As = ~Ais;

endmodule

ie BITWISE INVERT -> LOGICAL SHIFT LEFT -> BITWISE INVERT

This will work for any valid shift value.

http://www.edaplayground.com/x/YWK

Upvotes: 3

Morgan

Reputation: 20514

To have the shift work with up to the number of bits the following example pre pads the input with 1's shifts then selects the MSBs:

localparam WIDTH = 4;
// Create temp with 1's as padding
wire [WIDTH*2 -1 :0] pad   = {A, {4{1'b1}}};
wire [WIDTH*2 -1 :0] shift = pad << 1;

// Select MSB with 1's shifted in
wire [WIDTH-1 : 0] result = shift[WIDTH*2 -1 : WIDTH];

Upvotes: 0