Error in fi divide for embedded fi object
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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
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
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
Life is Wonderful
on 14 Oct 2021
Tom Bryan
on 14 Oct 2021
I have a few follow up questions to know how to answer:
- Are you trying to do code generation? This error only occurs during code generation.
- You have defined the variable power to be floating-point double. Do you want the code to be purely fixed-point?
- Do you want to generate C code? or a MEX function to simulate in MATLAB?
Best wishes,
Tom Bryan
Life is Wonderful
on 15 Oct 2021
Edited: Life is Wonderful
on 15 Oct 2021
Tom Bryan
on 15 Oct 2021
I'm using the methods described in article Best Practices for Converting MATLAB Code to Fixed Point. I'm the person who created the fi object and the "cast-like" syntax described there. By using this method, your code can run both fixed-point and floating-point.
A few notes about the code listed below:
- The range of the magnitude of the output is between
and 
. This means that you need at least 30 integer bits, and 30 fractional bits in the output. The output is also positive and negative. To accomodate this, I chose a signed fixed-point type with a 64 bit word length and 32 bit fraction length. - You can see in the generated code that the large word length of the output means that multiple words of builtin C types were needed for intermediate products and sums.
- I changed the names of the variables from base, exp, power to b, e, p respectively so the builtin functions of the same name would not be overloaded.
- I separated out the functions that you want to generate code for from the test bench.
- The test bench compiles and runs fixed-point as a MATLAB Executable (MEX).
- The test bench also generates C code for the fixed-point function.
- Run MATLAB script test_bench to test and compile the code.
- Save each of the functions enclosed below in their own file.
Best wishes,
Tom Bryan
Tom Bryan
on 15 Oct 2021
%% test_bench
%% Define input data
b = fi([-1:-1:-(2^3),1:1:(2^3)],1,32,27);
e = fi([-(2^4-6):1:-1,1:1:(2^4-6)],1,32,27);
%% Generate a MEX function for fixed-point inputs
codegen power_scalar_function -args {b(1),e(1)} ...
-o power_scalar_function_fixedpoint_mex -config:mex
%% Run fixed point
fprintf('\n---------------------------------\n')
fprintf('Fixed Point\n')
power_fixedpoint = power_loop(b,e);
%% Run floating point
fprintf('\n---------------------------------\n')
fprintf('Double\n')
power_float = power_loop(double(b),double(e));
%% Compare fixed point to floating point
fprintf('\n---------------------------------\n')
fprintf('Difference\n')
difference = power_float - double(power_fixedpoint) %#ok<NOPTS>
max_difference = max(abs(difference(:))) %#ok<NOPTS>
%% Generate C code
codegen power_scalar_function -args {b(1),e(1)} -config:lib -launchreport
%% Look at the stored integer values of power_fixedpoint
stored_integer_values = storedInteger(power_fixedpoint) %#ok<NOPTS>
Tom Bryan
on 15 Oct 2021
function p = power_loop(b,e)
T = power_types(b);
p = ones(length(b),length(e),'like',T.p);
fprintf('%20s|%20s|%20s|%20s|%20s|%20s|\n','i','j','power','base','exponent','NBits')
for i = 1:length(b)
for j = 1:length(e)
if isfi(b)
p(i,j) = power_scalar_function_fixedpoint_mex(b(i),e(j));
else
p(i,j) = power_scalar_function(b(i),e(j));
end
Bits = p(i,j);
NBits = log2(double(Bits));
fprintf('%20d|%20d|%20f|%20f|%20f|%20f|\n',i,j,p(i,j),b(i),e(j),NBits);
end
end
end
Tom Bryan
on 15 Oct 2021
function p = power_scalar_function(b,e)
T = power_types(b);
p = cast(1,'like',T.p);
if (e > 0)
while(e~=0)
p(:) = p * b;
e = accumneg(e, 1);
end
else
while(e~=0)
p(:) = p * 1/b;
e = accumpos(e , 1);
end
end
end
Tom Bryan
on 15 Oct 2021
function T = power_types(prototype)
if isfi(prototype)
% The range of the magnitude of the output is between 2^30 and
% 2^-30. This means that you need at least 30 integer bits, and 30
% fractional bits in the output. The output is also positive and
% negative. To accomodate this, I chose a signed fixed-point type
% with a 64 bit word length and 32 bit fraction length.
T.p = fi([],1,64,32);
else
% If not fixed-point, cast the output to the same type as the
% input.
T.p = cast([],'like',prototype);
end
end
Life is Wonderful
on 16 Oct 2021
Edited: Life is Wonderful
on 16 Oct 2021
Tom Bryan
on 16 Oct 2021
Thanks. It's my pleasure.
Could you provide a reproduction step to show me a bug in code generation for the power function?
For example, it would be a function using power that has a bug when you do codegen on it.
Best wishes,
Tom Bryan
Life is Wonderful
on 16 Oct 2021
Edited: Life is Wonderful
on 16 Oct 2021
Life is Wonderful
on 16 Oct 2021
Edited: Life is Wonderful
on 26 Oct 2021
Tom Bryan
on 16 Oct 2021
Can you tell me about the overall problem you're trying to solve? In what context are you using this function?
It may be possible to refactor the original problem into something that requires much less dynamic range. For example, it is often possible to work in the logarithmic domain instead of the exponential domain.
Best wishes,
Tom Bryan
tbryan@mathworks.com
Tom Bryan
on 16 Oct 2021
The error with the builtin fi power method is a limitation but not a bug. The error message is telling you how to set the fimath to meet the requirements to use with a fi object: "Exponent input to 'power' must be a real scalar and the value must be a non-negative integer." See the documentation for fimath for how to use it with a fi object.
After setting fimath, you will run into another limitation with the builtin fi power method. It only works with positive exponents, and your example uses positive and negative exponents. Again, this is a limitation but not a bug.
Life is Wonderful
on 17 Oct 2021
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