A problem with floating point arithmetic

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

Answer 1

[Matt Fig]

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

[Nick]

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!