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This simple code note working, HELP

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Poojitha Ariyathilaka
Poojitha Ariyathilaka el 13 de Jun. de 2020
Comentada: Poojitha Ariyathilaka el 13 de Jun. de 2020
%this works well upto t==0.7, but not after 0.8
%give me an answer for this, not links
t=0;
for n=1:10
t=t+0.1;
disp(t)
if t==0.8
disp('if')
else
disp('else')
end
end

Respuesta aceptada

Ameer Hamza
Ameer Hamza el 13 de Jun. de 2020
Editada: Ameer Hamza el 13 de Jun. de 2020
Direct comparison using == of floating-point number does not give the correct result. You need to allow some level of tolerance. Try this
t=0;
for n=1:10
t=t+0.1;
disp(t);
if abs(t-0.8) < 1e-6
disp('if')
else
disp('else')
end
end
  3 comentarios
Ameer Hamza
Ameer Hamza el 13 de Jun. de 2020
yes, that will work as well, but it is good to use the tolerance method. Otherwise, you will need to do rounding every time you do some mathematical operation on the variable.
Poojitha Ariyathilaka
Poojitha Ariyathilaka el 13 de Jun. de 2020
Yes, tolerance method is much quicker.

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