where is the error in this code?

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==1.8

disp('ok');

end

b=b+0.1;

end

Answer 1

[Vilém Frynta]

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

[Dyuman Joshi]

A better syntax would be

if abs(z-1.8)<tolerance

Where you can set tolerance according to the requirement.

[Herman Khalid]

Thank you for your answers, but I tested from 0.0 to 5.0, just 1.8 had this issue?

[Herman Khalid]

try to test with 1.6 or any other number you will get correct result

[Vilém Frynta]

@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]

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'

Notice where the values differ?

So, while dealing with floating point numbers, always use a tolerance to compare.

Answer 2

[Image Analyst]

See the FAQ for a thorough discussion:

https://matlab.fandom.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_(or_similar)_not_equal_to_zero?