Asked by Ali Tawfik
on 25 Aug 2019

I am doing a loop, and want the iteration stops when condition of x = 5 however, the code terminates before x=5 ?

So any help please

clear all;

clc;

x=[1 2 3 4 5 6 7 8 9 10];

for n=1:10

if x(n)>= 5;

break

end

disp('how many iteration done now');

final(n,:)=x(n)

z(n,:)=x(n)+2

end

Answer by dpb
on 25 Aug 2019

Accepted Answer

break is like drawing the "Go To Jail" card in Monopoly--you "go directly to jail, do not pass GO, do not collect $200". Here when you get to the conditional and it is satisfied, then the code immediately goes to the next line of code after the terminating end of the for loop and so the other code isn't executed that last iteration--the loop did do five iterations here because it took until then for x(n) to be >=5; just that you didn't do anything at all in the loop other than the test.

Perhaps your expectations could be met simply by reordering:

...

for n=1:10

disp('how many iteration done now');

final(n,:)=x(n)

z(n,:)=x(n)+2

if x(n)>= 5, break, end

end

...

Then again, that assumes that what happens now is what you want to happen for the loop for all iterations that it does do...

dpb
on 25 Aug 2019

You left off the second (and optional, but needed here) argument to find

Instead of me spoon-feeding, look at the documentation for find and see if you can't learn the difference between the two invocations yourself.

Going forward, it will be much more productive if you learn to read and use the doc instead of waiting for somebody here to answer a question that can be resolved thru the input descriptions.

Ali Tawfik
on 26 Aug 2019

I have already read the find doc, seems very helpful,

However, I don't agree with you in what you said about optional, or even your last code in general

(x>=5,1) 1 here means 1 which means first element that met the condition,

second, why you add the values to 2 ?

I meant, I need from x array all the numbers that met the conditions written.

Your obtained values are uncorrect.

Thanks,

dpb
on 26 Aug 2019

"Optional" in that it is an optional input to find(), not that it is optional to use it in this case to find the first location in x that satisfies the condition which is what your break logic is doing.

I added 2 because that's what your code does... :)

It produced identically the same results as your code here...

>> x=1:10;

>> for n=1:10

disp('how many iteration done now');

final(n,:)=x(n);

z(n,:)=x(n)+2;

if x(n)>= 5, break, end

end

how many iteration done now

how many iteration done now

how many iteration done now

how many iteration done now

how many iteration done now

>> [final z(:,5)]

ans =

1 3

2 4

3 5

4 6

5 7

>>

Your code above w/o the intermediate results while mine produced:

>> final=find(x>=5,1);

>> z=x(1:final)+2;

>> final

final =

5

>> z.'

ans =

3

4

5

6

7

>>

with the assumption as before that there really wasn't any need for final to be an array and the desired result was that for z when the condition is met.

Now, if that isn't the result wanted, then the actual problem definition needs clarification. So, if the problem is really the statement "I meant, I need from x array all the numbers that met the conditions written." then that's just

z=x(x>=5);

using logical addressing/indexing, one of the most powerful of Matlab features. That isn't what your code actually does, but maybe that was the real question underneath the one you asked.

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## Rik (view profile)

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https://www.mathworks.com/matlabcentral/answers/477521-getting-values-from-a-for-loop-when-specific-condition-met#comment_738848

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