How to use coder.ceval to call a void type C++ function

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Erjian
Erjian 2021-4-2
评论： Darshan Ramakant Bhat 2021-4-5
Hi,
I am trying to call an external C++ function, which is a void type function. Here is the code I wrote as an example
function [Y] = example(a,b)
coder.updateBuildInfo('addSourceFiles','cpp_fun.cpp');
coder.cinclude('cpp_fun.h');
coder.ceval('cpp_fun', a, b, Y);
end
The C++ function is defined as:
void cpp_fun(int a, float *b, float *Y);
Here int a, float *b are input of the function and *Y is suppose to be the output of the C++ function.
But when I use the Matlab Coder to convert the matlab function above, I get the following error message: "Variable 'Y' is not fully defined on some execution paths."
Shoud I use something like
Y = coder.opaque(type)
in the matlab function to define Y? If yes, what type should I use for Y?
Any help is greatly appreciated!

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回答（1 个）

Darshan Ramakant Bhat
Try defining the variable 'y' in MATLAB function then try to use coder.wref()
Y = 0;
coder.ceval('cpp_fun', a, b, coder.wref(Y));
https://www.mathworks.com/help/simulink/slref/coder.wref.html
Similarly you may need to use coder.ref() for the second input 'b' since it is a pointer.
https://www.mathworks.com/help/simulink/slref/coder.ref.html
Hope this will help you.
  2 个评论
Erjian
Erjian 2021-4-2
Hi Darshan,
Thanks fo your help! While I was trying your suggestions and I ran into other problems. I simplifed my C++ function to better understand how this should work. Here is the modified matlab function:
function [Y] = example(a,b)
if coder.target('MATLAB')
% Executing in MATLAB, call function example
Y = a+b;
else
coder.updateBuildInfo('addSourceFiles','cpp_fun.cpp');
coder.cinclude('cpp_fun.h');
Y = 0;
coder.ceval('cpp_fun', a, b, coder.wref(Y));
end
end
And here is the simplified C++ code:
void cpp_fun(int a, int b, float y)
{
y = (float)a + (float)b;
}
And here is the screeshot of the error I got in the Matlab Coder Report Viewer:
How should I handle this data type inconsistency problem?
Darshan Ramakant Bhat
Darshan Ramakant Bhat 2021-4-5
You need to use proper types at the interface. From the error I can see that you are passing "y" as pointer but in C++ code you are taking by value. This is giving the error. Also type of "y" is real_T, you need to make it as float.
I have modified your example and made it work. Please check the attached file and run doit.m file.
Hope this will help you.

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