I have trouble understanding the command.

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I'm having trouble understanding what the command below does, I'd like to know if anyone knows what it does.
for i=1:nset
tty=y(i,:);
index=find(tty>=0);
tx=x(i,index,:);
ty=tty(index);
tpar=mpar(:,i);
command=['[esty seout]=',fun,'(tpar,tx);']; //
eval(command); 
m=length(esty);
predy(i,1:m)=esty;
if(linear == 1);
errory=errory+((esty-ty)*(esty-ty)');
else
findex=find(esty<=0);
esty(findex)=ones(size(findex))/1000000000;
error=20*log10(esty)-20*log10(ty);
errory = errory+error*error';
end
end
I don't know what you are doing with the two sentences in this programme, does anyone know?
command=['[esty seout]=',fun,'(tpar,tx);'];
eval(command); 
  2 Comments
Stephen23
Stephen23 on 17 Oct 2022
Edited: Stephen23 on 17 Oct 2022
Do not learn from such badly-written code.
If the code's author had read the Tips section of the EVAL documentation, they would know it is not recommended to include the output arguments in the EVAL call, only the RHS. They could (should) have done that.
And then if they had done a bit more thinking and reading on the topic, they would have known to replace the EVAL with STR2FUNC (which defines a function handle that can be repeatedly called without the performance hit of repeated EVAL calls).
Caecus caeco dux. Sigh.

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Accepted Answer

Hiro Yoshino
Hiro Yoshino on 15 Oct 2022
You should have a good understanding on eval by going through this.
What it does here is that
[esty, seout] = fun(tpar,tx);
It seems there is a function named "fun" somewhere and its arguments are tpar and tx.
  5 Comments
Hiro Yoshino
Hiro Yoshino on 22 Oct 2022
@Bruno Luong You're right. That was my fault. Sorry guys but it's good to know that you found the "fun" function somewhere.

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