shift - helpneeded with codeexplanation

[jasonkee111]

shift

can somebody explain the code below

input wire signed [16:0] D,
output reg signed [16:0] Q );

always @ * begin
Q = D;
for(j=0;j<i;j=j+1)
Q = (Q >> 1) | (D[16] << 16);

Question:
1. What is the meaning D[`16] << 16? from i understand, it shift left 16 times of D whereby D only 1 bit due to D[16]. Am i misunderstand?

2. why it use Q and D in the statement "Q = (Q >> 1) | (D[16] << 16);" since D is assigned to Q?

Thanks

 

[ckaa]

Re: shift

I am not completely sure as to why this happens but it looks like (D[16] << 16) takes your sign bit and and assigns it to Q[16] . Since both your variables are signed, when Q is shifted by one, the MSB bit is maintained by 'ORing' it with (Q>>1). I'd be interested to know details about this if you find out more!