This simple code note working, HELP

2020 6 13
1 个回答
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更新时间：2020 6 13
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%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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 采纳的回答

Ameer Hamza
Ameer Hamza 2020-6-13
编辑：Ameer Hamza 2020-6-13

1 个投票

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 个评论
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Thank you very much, that worked!
If the case comprises with the floating point, can't we do it by rounding?
t=0;
for n=1:10
t=round(t+0.1,1);
disp(t)
if t==0.8
disp('if')
else
disp('else')
end
end
Ameer Hamza
Ameer Hamza 2020-6-13
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.
Yes, tolerance method is much quicker.

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