What is the type of A, B, CIN and RESULT?
The leading '0' might be used to set the MSB =sign bit.
Dear Prasad
Thank's in this question the A and B are N bits and CIN is one bit and RESULT is N+1 bits
Hi Farzin Piltan,
I have synthesized the following piece of code, which looks up to your logic.
library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
use IEEE.STD_LOGIC_ARITH.ALL;
use IEEE.STD_LOGIC_UNSIGNED.ALL;
entity Adder is
port (A,B: in std_logic_vector(3 downto 0);
CIN: in std_logic;
RESULT1: out std_logic_vector(4 downto 0);
RESULT2: out std_logic_vector(5 downto 0));
end Adder;
architecture Behavioral of Adder is
begin
RESULT1
Hi Farzin Piltan,
The inference from the above synthesis is that the widths have to be appropriately set for the signals. In the second case, the case of RESULT2, the reason for the assigned expression to be of 9-bit wide is that the sum of the signals, with concatenated values, assumes the sum of their bit-widths as the bit-width of the obtained sum. To correct this, assign the concatenated signals as some other intermediate signals, and then add them.
in digital electronics, for a Full adder, all inputs must be of same length
Dear Prasad
Thank you, but how I can explain the difference between these two methods and when we used the first and when we used the second one????
Hi Farzin Piltan,
You are welcome! Logically, there is no difference between the two methods except for their bit-widths. So, you may use any of the two anywhere, provided you make a check for their sizes.
However, if this computational unit (adder in both the cases) is a part of another design, then depending on the required size of the data, proper expression has to be chosen.
The first statement avoids overflow by increasing the size of the result by one bit. In the second statement overflow is possible.
Example: N=3, A = 111, B = 000, CIN = 1
1) RESULT = 1000 (no overflow)
2) RESULT = 000 (overflow)
To N. Prasad:
The expression you wrote for RESULT2 in your code above contains an error. Since you forgot the brackets combining ('0'&A) and ('0'&B), your expression turned into
RESULT2
To Farzin Piltan:
To summarize the explanations given by my colleagues above :)
The difference between
RES1
Hi Farzin,
As mentioned by Mario, the second statement overflows and will cause errors during compilation if the size of the output and inputs are not same.
For the time being forget about CIN.
So your code is,
(1) RESULT
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