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Altera_Forum
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Honored ContributorFloating point addition in vhdl
Hii
I want to add 4 floating point no.what will be the result range.I am using library ieee_proposed; use ieee_proposed.fixed_pkg.all; this package.For example if I want to add two floating point no then the result range is C= A+ B -- range of C is (max(A'right ,B'right)+1 downto min (A'left ,B'left)) and it is working fine .But when I am doing C = A+B+D+E; What will be the new C range. If I am following the same as above I am getting an error "D:/214ee1411/vhdl codes/floatarith/mul_1test/fmul_1.vhd" Line 50: Expression has 11 elements ; expected 9 Please help me with this. Thank you11 Replies
Altera_ForumYou are using fixed point?
use ieee_proposed.fixed_pkg.all;- Altera_Forum--- Quote Start --- You are using fixed point? use ieee_proposed.fixed_pkg.all; --- Quote End --- Thank you for your reply. Yes I am using fixed point.For example I want to add s1 <=((to_ufixed (50,7,0))+(to_ufixed (50,7,0))+(to_ufixed (50,7,0))+ (to_ufixed (50,7,0))); Here is there any formulation to calculate the range of s1?
- Altera_Forum
Yes,
Please see table at http://electro-logic.blogspot.it/2015/11/fpga-numeri-virgola-fissa-seconda-parte.html where there are sizing rules - Altera_Forum--- Quote Start --- Yes, Please see table at http://electro-logic.blogspot.it/2015/11/fpga-numeri-virgola-fissa-seconda-parte.html where there are sizing rules --- Quote End --- Thank you for the link.I am having this table.Here it is given for A and B only.My question is for adding 4 times..as if I am taking the s1 range (8 downto 0) as per table.It is showing error which I have posted above. When I an going for s1 range (10 downto 0 ) it is working fine.So my question is how to decide this range if I want to add multiple times.
- Altera_Forum
Just break it down in to multiple 2 input additions. There are 3 additions, so the output needs to be 3 bits bigger than the largest input.
- Altera_Forum--- Quote Start --- Just break it down in to multiple 2 input additions. There are 3 additions, so the output needs to be 3 bits bigger than the largest input. --- Quote End --- Thank you all
- Altera_Forum--- Quote Start --- Just break it down in to multiple 2 input additions. There are 3 additions, so the output needs to be 3 bits bigger than the largest input. --- Quote End --- Excuse me for being a bit of a bore, but if you add up 4 values, an extension with 2 bits suffices ...
- Altera_Forum--- Quote Start --- Excuse me for being a bit of a bore, but if you add up 4 values, an extension with 2 bits suffices ... --- Quote End --- Algorithmically yes - but not syntactically. Each addition in the fixed point library increases the bit size.
- Altera_Forum--- Quote Start --- Algorithmically yes - but not syntactically. Each addition in the fixed point library increases the bit size. --- Quote End --- I haven't used fixed-point types yet. But if we do it as you said: break it down into multiple 2 input additions:
2 bits extra will suffice.result := (a + b) + (c + d) - Altera_Forum
Yes, now you've added brackets, it will only need 2 extra bits.
But a+b+c+d will break down into:
which does require 3 extra bits.result := (((a+b) + c) + d);
