CODEWITHSUNDEEP
hardwareverilogfpgahdl
ipunished
ipunished

Reputation: 694

else block of if-else acting differently to different conditions

I'm coding a simple shift register using if else block. I noticed that the else block works as it should when the control signal is control = 2'b00 (meaning it retains the default vale) but when I give the control value control = 2'b11 it starts shifting to the right, which is not what I want.

Why does the else block work selectively? Even when both control = 2'b00 and control = 2'b11 fall in the else case?

Code and screenshot below:

module shift(
input clock,
input reset,
input [1:0] control,
input in,
output [7:0] out
);

reg [7:0] r_reg, r_next; //a 7 bit shift register which will be output as is, this can be changed to any size

always @ (posedge clock or posedge reset)
begin
    if(reset)
        r_reg <= 0;
    else
        r_reg <= r_next;
end

always @ (*)

begin

if(control[0]) //shift right
    r_next = {in, r_reg[7:1]};

else if(control[1]) //shift left
    r_next = {r_reg[6:0], in};

else
    r_next = r_reg; //default state stays the same

end

assign out = r_reg;

endmodule

enter image description here

EDIT: 

if(right) //shift right
    r_next = {in, r_reg[7:1]};

else if(left) //shift left
    r_next = {r_reg[6:0], in};

else if((~right & ~left) || (right & left))
    r_next = r_reg; //default state stays the same

The above did not work either.. But I fixed it with case.

case(control)
    2'b01: r_next = {in, r_reg[7:1]};
    2'b10: r_next = {r_reg[6:0], in};
    default: r_next = r_reg;

Upvotes: 0

Views: 124

Answers (2)

Carl Norum
Carl Norum

Reputation: 224972

control[0] is 1 when you input 11, so I think it's working as expected. You'd need a stronger condition to make it work the way you seem to want:

if (control == 2'b01)          // shift right
    r_next = {in, r_reg[7:1]};

else if (control == 2'b10)     // shift left
    r_next = {r_reg[6:0], in};

else                           // default state stays the same
    r_next = r_reg;

Upvotes: 2

Ben Voigt
Ben Voigt

Reputation: 283713

Very simple: It doesn't fall in the else case at all.

Your first condition looks only at the low bit (matches b'd1), therefore both b'01 and b'11 shift right.

Upvotes: 2

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