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where is the error in this code?

조회 수: 1 (최근 30일)
Herman Khalid
Herman Khalid 2023년 4월 22일
답변: Image Analyst 2023년 4월 22일
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

답변 (2개)

Vilém Frynta
Vilém Frynta 2023년 4월 22일
편집: Vilém Frynta 2023년 4월 22일
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
  댓글 수: 5
Vilém Frynta
Vilém Frynta 2023년 4월 22일
@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
Dyuman Joshi 2023년 4월 22일
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.

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Image Analyst
Image Analyst 2023년 4월 22일

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