Effacer les filtres
Effacer les filtres

Error observed in using bitsll and bitsra

3 vues (au cours des 30 derniers jours)
Life is Wonderful
Life is Wonderful le 22 Sep 2022
When utilizing bitsll and bitsra, an error was noticed.
I have the following code and intend to utilize Embedded.fi [fixed point ] as double implementation show in case -1.
Can someone please show me how to code correctly and help me understand what I'm missing in cases 2 and 3?
It would be nice if I could do cases 2 and 3 in a single function/ single solution.
Thank you very much.
% case -1 : double
k = linspace(-12,12,10);
for i = 1 : length(k)
a = 2 ^ k(i)
a = 2.4414e-04
a = 0.0016
a = 0.0098
a = 0.0625
a = 0.3969
a = 2.5198
a = 16
a = 101.5937
a = 645.0796
a = 4096
% case -2: Embdedded fi ,fixed point - right shift
k = fi(linspace(-12,12,10),1,32,27);
a = int8(1);
for i = 1 : length(k)
% fxpOut = bitsra(a,k(i))
Error using bitsra
K must be numeric, scalar, real, integer-valued, and greater than or equal to zero in BITSRA(A,K).

Error in embedded.fi/bitsra (line 38)
y = bitsra(x, double(kin));
% case - 3: Embdedded fi ,fixed point - left shift
k = fi(linspace(-12,12,10),1,32,27);
a = int8(1);
for i = 1 : length(k)
fxpOut = bitsll(a,k(i))
  4 commentaires
Richard McCormack
Richard McCormack le 26 Sep 2022
I am happy that you made progress!
I made some adjustments to your code to make it work for code generation.
But maybe there is a reason that you can't use abs. If that is the case, please let me know why you can't use abs.
function scratch
K = [-12,12];
for i = 1:length(K)
if sign(K(i)) < 0 % Bitsliceget to know bitposition and do a bitand for sign
% Negative
b = fi(bitsra(1,abs(K(i))));
% Positive
b_i = fi(bitsra(1,abs(K(i))));
Life is Wonderful
Life is Wonderful le 26 Sep 2022
Modifié(e) : Life is Wonderful le 28 Sep 2022
Thanks a lot @Richard McCormack.
I am fine in using abs. I will try out your suggested code and make some test.

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