I code a bisection method on MATLAB app designer, how can I fix this error ?

16 vues (au cours des 30 derniers jours)
Anne
Anne le 30 Nov 2023
typefunction = app.TypefunctionEditField.Value;
a = app.aEditField.Value
b = app.bEditField.Value
run = app.RunButton.ButtonPushedFcn
error= app.errorEditField.Value
fxi=str2func(['@(x)'typefunction]);
number_of_iterations=0;
if fxi(a)*fxi(b)>0
result=NaN;
return
end
if fxi(a)==0
result=a;
return
elseif fxi(b)==0
result=b;
return
end
while(abs(a-b)>=error)
result=(a+b)/2;
if fxi(a)*fxi(result)<0
b=result;
else
a=result;
end
number_of_iterations=number_of_iterations+1;
end

Réponses (1)

Walter Roberson
Walter Roberson le 30 Nov 2023
fxi=str2func(['@(x)'typefunction]);
You need a space or comma, like
fxi=str2func(['@(x)' typefunction]);
  2 commentaires
Anne
Anne le 30 Nov 2023
i fix it but it still have same error :(
Walter Roberson
Walter Roberson le 30 Nov 2023
I copied your posted code into the editor, and added in the one space that I indicate, and that cleared up the error message. There are still warnings recommending that you add in semi-colons to prevent unnecessary output, but the error is gone just by putting in that one space.

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