[Fixed Point Converter] 1000 bit fraction length at 2^x if x is loaded from .mat file?

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I got a problem with code like this in the fixed-point converter:
dut_tb.m
dut(8);
dut.m
function out = dut(in)
S = coder.load('constants.mat');
out = in+2^S.exponent;
end
storeConstants.m
function [] = storeConstants()
exponent = 8;
save('constants.mat');
end
convert.m
exArgs = {8};
fc = coder.config('fixpt');
fc.TestBenchName = 'dut_tb';
codegen -float2fixed fc -args exArgs dut
Reproduction steps:
run storeConstants and then convert.
The fixed-point converter automatically uses a fraction length of 1000 for the power of two operation.
As this is just a simple bitshift in hardware, I cannot see the purpose of this high precision. How can
I disable this huge fraction length?

Respuesta aceptada

Jan Siegmund
Jan Siegmund el 15 de Abr. de 2020
A temporary fix is using bitsll as recommended here:
For efficient HDL code generation, use the Fixed-Point Designer functions bitsll, bitsrl, or bitsra instead of bitshift.
So i exchanged
out = in+2^S.exponent;
to
out = in+bitsll(uint8(1),S.exponent);
Uint8 being used, because I want the result to be unsigned, because it needs to be used in a divider.
  1 comentario
Kiran Kintali
Kiran Kintali el 18 de Abr. de 2020
Thank you for the inputs. We will see if we can generate this recommendation during the automatic conversion process in a future release.

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Más respuestas (1)

zhang zheng
zhang zheng el 13 de Abr. de 2020
i also sometimes find fixedpoint-tool is not easy to understand. i often first convert m-code from float point to fixed point, and then convert them to hdl code or c code.
  2 comentarios
Jan Siegmund
Jan Siegmund el 15 de Abr. de 2020
So you mean converting to fixed-point, then editing the bad parts of the fixed-point code by hand and then converting to code?
But this 1000 fraction length only occurs, if the exponent is loaded from a .mat file, so I expect it to be a bug.
Jan Siegmund
Jan Siegmund el 15 de Abr. de 2020
Ok I got a temporary fix. See my answer for details.

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