where is the error in this code?
显示 更早的评论
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
回答（2 个）
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
5 个评论
Dyuman Joshi
2023-4-22
A better syntax would be
if abs(z-1.8)<tolerance
Where you can set tolerance according to the requirement.
Herman Khalid
2023-4-22
Herman Khalid
2023-4-22
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.
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)
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
Notice where the values differ?
So, while dealing with floating point numbers, always use a tolerance to compare.
Image Analyst
2023-4-22
0 个投票
See the FAQ for a thorough discussion:
类别
在 帮助中心 和 File Exchange 中查找有关 Logical 的更多信息
2023-4-22
2023-4-22
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
