This simple code note working, HELP

조회 수: 1 (최근 30일)
Poojitha Ariyathilaka
Poojitha Ariyathilaka 2020년 6월 13일
댓글: Poojitha Ariyathilaka 2020년 6월 13일
%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

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Ameer Hamza
Ameer Hamza 2020년 6월 13일
편집: Ameer Hamza 2020년 6월 13일
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
  댓글 수: 3
Ameer Hamza
Ameer Hamza 2020년 6월 13일
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 2020년 6월 13일
Yes, tolerance method is much quicker.

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