[HDL Coder] Bug: Fixed point converter does not descend into System Objects

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Can I somehow tell the fixed point converter to explicitly convert a certain function? This would temporarily fix the following bug.
The Bug I found can be reproduced with these files:
dut_tb.m
a = dut(1);
dut.m
function out = dut(in)
persistent system;
if isempty(system)
system = dutSystem(1.1);
end
out = system.step(in);
end
dutSystem.m
classdef dutSystem < matlab.System
properties (Nontunable)
number
end
methods (Access = public)
function obj = dutSystem(number)
obj.number = number;
end
end
methods (Access = protected)
function out = stepImpl(obj,in)
out = buggy_function(in,obj.number);
end
end
end
buggy_function.m
function out = buggy_function(in1, in2)
out = round(in1 + in2);
end
convert.m
function [] = convert()
exArgs = {1};
fc = coder.config('fixpt');
fc.TestBenchName = 'dut_tb';
hc = coder.config('hdl');
codegen -float2fixed fc -args exArgs -config hc dut
end
Execute convert. The HDL coder will stop, because round is supposedly not supported by the HDL coder. But it is supported. The fixed-point converter just did not convert buggy_function.m

Respuestas (1)

Kiran Kintali
Kiran Kintali el 20 de Abr. de 2020
Unfortunately the development team cannot reproduce the issue. Is it possible the attached reproduction steps are not sufficient? Can you share a zip file with the code and a snapshot of the error message when running the convert() function? Thanks
  1 comentario
Jan Siegmund
Jan Siegmund el 22 de Abr. de 2020
Editada: Jan Siegmund el 22 de Abr. de 2020
Sorry, somehow, MATLAB 2020a in contrast to 2019b accepts double inputs in round.
But this does not mean, that the error is not reproduceable. The problematic extract from my dutSystem_fixpt.m:
...
function out = stepImpl(obj,in)
fm = get_fimath();
out = fi(buggy_function(in,obj.number), 0, 14, 15, fm);
end
...
It uses the standard buggy_function and not buggy_function_fixpt.
Use this buggy_function.m instead and you will see an error pop up.
function out = buggy_function(in1, in2)
out = in1/(2^in2);
end

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