A problem with floating point arithmetic

2011 6 7
1 个回答
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Respond=[0.000065;0.000032;0.000120;0.000231];
for i=0.000000:0.000001:0.000350
for k=1:4
if Respond(k,1)==i
fprintf('i: %d\n',i);
fprintf('k: %d\n',k);
end
end
end
Problem: it prints only the 2 of the 4 slots of table Respond :S Any idea?
Thanks

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 采纳的回答

Matt Fig
Matt Fig 2011-6-7

0 个投票

You are seeing the limits of floating point arithmetic. This is why it is not recommended to use the equality operator on floating point numbers....
What is it you actually want to do??
Here is the difference:
Respond=[0.000065;0.000032;0.000120;0.000231];
for i=0:0.000001:0.000232
for k=1:4
if abs(Respond(k)-i)<20*eps
fprintf('i: %20.20f\n',i);
fprintf('Respond(k): %20.20f\n',Respond(k))
end
end
end

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Nick
Nick 2011-6-7
I am doing a simulation about a network and working with 10^-7 timeslots..But i think that i will use higher numbers.Insted of 0.000001 i will use 1.
Thnx a lot for the answer!

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