where is the error in this code?

In computer programming, when dealing with floating-point numbers, there can be slight differences in how these numbers are represented due to the finite number of bits used. As a result, when comparing such numbers using the "==" operator, unexpected results may occur due to these small errors in their representation.

Someone correct me if I'm wrong.

clc,clear,close
v=[1.8 1.8 1.8 1.6];
a=1;
b=0.0;
for j=1:9
    z=a+b;
    if z >= 1.799999999 && z <= 1.800000001
        disp('ok');
    end
    b=b+0.1;
end
ok

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Vilém Frynta 2023-4-22

@Dyuman Joshi thanks for your addition. i just tried to quickly show what i had on my mind.

@Herman Khalid it's possible. i do not have any deeper understanding this topic, but perhaps you could say that it's 'random'.

also, if you found my answer helpful, i'd be glad if you could accept it.

Dyuman Joshi 2023-4-22

在 MATLAB Online 中打开

Some floating point numbers can not be represented exactly in binary form.

You can see below what the values are stored as -

v = [1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9];
vs = sprintf('%20.18f\n',v)
vs = 
    '1.000000000000000000
     1.100000000000000089
     1.199999999999999956
     1.300000000000000044
     1.399999999999999911
     1.500000000000000000
     1.600000000000000089
     1.699999999999999956
     1.800000000000000044
     1.899999999999999911
     '

Now compare that to the values of z -

a=1;b=0.0;
format long
for j=1:10
    z=a+b;
    zvalue=sprintf('%20.18f',z)
    if z==1.8
        disp('ok');
    end
    b=b+0.1;
end
zvalue = '1.000000000000000000'
zvalue = '1.100000000000000089'
zvalue = '1.199999999999999956'
zvalue = '1.300000000000000044'
zvalue = '1.399999999999999911'
zvalue = '1.500000000000000000'
zvalue = '1.600000000000000089'
zvalue = '1.699999999999999956'
zvalue = '1.799999999999999822'
zvalue = '1.899999999999999911'