Error in fi divide for embedded fi object

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Jogger
Jogger on 14 Oct 2021
Edited: Jogger on 26 Oct 2021
Hi
I am trying to make a fixed point calculation with below code. But I run into error , I don't understand what is wrong in my code.
Note : Error results
clearvars;clc;
% define ranage for input data
base = fi([-1:-1:-(2^3),1:1:(2^3)],1,32,27);
exp = fi([-(2^4-6):1:-1,1:1:(2^4-6)],1,32,27);
power = 1;
if (exp > 0)
while(exp~=0)
power = power * base;
exp = accumneg(exp, 1);
end
else
while(exp~=0)
power = power * 1/base;
exp = accumpos(exp , 1);
end
end
Error using embedded.fi/mrdivide>fidivide (line 56)
For fi objects, B must be a scalar in A/B.

Error in / (line 23)
c = fidivide(a,b);
Without fi object code, below code is fine
clearvars;clc;
% define ranage for input data
base = [-1:-1:-(2^3),1:1:(2^3)];
exp = [-(2^4-6):1:-1,1:1:(2^4-6)];
power = 1;
fprintf('%20s|%20s|%20s|%20s|%20s|%20s|\n','i','j','power','base','exponent','NBits')
for i = 1:length(base)
for j = 1:length(exp)
power(i,j) = power_flpt(base(i),exp(j));
Bits = power(i,j);
NBits = log2(Bits);
fprintf('%20d|%20d|%20f|%20f|%20f|%20f|\n',i,j,power(i,j),base(i),exp(j),NBits);
end
end
function power = power_flpt(base,exp)
power = 1;
if (exp > 0)
while(exp~=0)
power = power * base;
exp = accumneg(exp, 1);
end
else
while(exp~=0)
power = power * 1/base;
exp = accumpos(exp , 1);
end
end
end

Accepted Answer

Tom Bryan
Tom Bryan on 14 Oct 2021
The error is telling you that 1/base isn't supported in fi because base is a vector. The error is telling you that base must be a scalar for mrdivide (matrix right-divide). Element-wise division is 1./base, which would not have thrown an error, but the first code was logically incorrect anyway. The first code didn't give the right answer with double either. It was using things like "if (exp > 0)" where exp is a vector, so the if condition was never entered. Also the "while(exp~=0)" condition was never entered either.
The second code works with both fixed-point and floating-point becasue the power_fltpt function only uses scalars.
Also, both exp and power are builtin MATLAB functions, so you should think about renaming them. MATLAB is forgiving about overloading other functions, but it may interfere if you ever want to use the exp or power functions.
Enclosed below is your second code running with both fixed-point and floating point, and comparing the answers at the end.
clearvars;clc;
% define ranage for input data
base = fi([-1:-1:-(2^3),1:1:(2^3)],1,32,27);
exp = fi([-(2^4-6):1:-1,1:1:(2^4-6)],1,32,27);
fprintf('\n---------------------------------\n')
fprintf('Fixed Point\n')
power_fixedpoint = power_loop(base,exp);
fprintf('\n---------------------------------\n')
fprintf('Double\n')
power_float = power_loop(double(base),double(exp));
fprintf('\n---------------------------------\n')
fprintf('Difference\n')
difference = power_float - double(power_fixedpoint) %#ok<NOPTS>
max_difference = max(abs(difference(:))) %#ok<NOPTS>
function power = power_loop(base,exp)
power = ones(length(base),length(exp));
fprintf('%20s|%20s|%20s|%20s|%20s|%20s|\n','i','j','power','base','exponent','NBits')
for i = 1:length(base)
for j = 1:length(exp)
power(i,j) = power_scalar_function(base(i),exp(j));
Bits = power(i,j);
NBits = log2(Bits);
fprintf('%20d|%20d|%20f|%20f|%20f|%20f|\n',i,j,power(i,j),base(i),exp(j),NBits);
end
end
end
function power = power_scalar_function(base,exp)
power = 1;
if (exp > 0)
while(exp~=0)
power(:) = power * base;
exp = accumneg(exp, 1);
end
else
while(exp~=0)
power(:) = power * 1/base;
exp = accumpos(exp , 1);
end
end
end
Best wishes,
Tom Bryan
  15 Comments
Jogger
Jogger on 17 Oct 2021
Many thanks . Yes , I can understand limitation !!
best regards

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