store LSB after shift right - vhdl

[fanwel]

Hi,

In my project I need to right shift an input vector by one. For example I have A=0101. Then right shift by one I get Q=0010. Now, I want to store the LSB of the input vector in this case is bit '1'. How to do the operation for store the LSB value? There is no problem to write vhdl code for the right shift operation, but I don't know how to write for the store LSB operation :-(. Can anyone give idea or advice. Many thanks

 

[alexan_e]

You mean how to access individual bits of a vector?
You can use A(0) or A(1) or A(1 downto 0) etc

 

[fanwel]

You mean how to access individual bits of a vector?
You can use A(0) or A(1) or A(1 downto 0) etc

Hi alexan_e,

Actually I want to store the LSB until reach 4 values (from different input vector). Then the output will be a vector of (3 downto 0). Is it possible? If yes, how can I do that? Thanks

 

[alexan_e]

The concatenation operator is & so you can use

Q<= A(0) & B(0) & C(0) & D(0);

 

[fanwel]

The concatenation operator is & so you can use

Q<= A(0) & B(0) & C(0) & D(0);

Hi alexan,

I understand your above idea. But, the input is of std_logic data type, whilst the output is of std_logic_vector data type. Many thanks for your response

 

[alexan_e]

If the input (A,B,C,D) is std_logic then just use

Q<= A & B & C & D;

I assume Q is 4 bit wide, if not you can assign a sub range using

Q(3 downto 0)<= A & B & C & D;

 

[fanwel]

If the input (A,B,C,D) is std_logic then just use

Q<= A & B & C & D;

I assume Q is 4 bit wide, if not you can assign a sub range using

Q(3 downto 0)<= A & B & C & D;

Hi alexan_e,

Yes, I understand that. How to wait the input to reach four value from one value before the output with four bit width is come out.I attach the figure for the component.Many thanks for response

 

[alexan_e]

You can use a state machine of some kind that assign the vector bits, one at a time each time there is a shift

something like
first step QLSB(3)<= Q(0)
second step QLSB(2)<= Q(0)
third step QLSB(1)<= Q(0)
fourth step QLSB(0)<= Q(0)

where QLSB will be the resulting vector

 

[fanwel]

You can use a state machine of some kind that assign the vector bits, one at a time each time there is a shift

something like
first step QLSB(3)<= Q(0)
second step QLSB(2)<= Q(0)
third step QLSB(1)<= Q(0)
fourth step QLSB(0)<= Q(0)

where QLSB will be the resulting vector

Hi alexan_e,

Thanks for give an idea. I will try the code. Many thanks