use of "break"

Hi everyone,

I have a few lines of code that look smth like this

exist = false;

for j = 1:1:J

if phi(m,j-1,n)*phi(m,j,n)<=0

x0 = SL+(j-2)*dS;

x1 = SL+(j-1)*dS;

y0 = phi(m,j-1,n);

y1 = phi(m,j,n);

boundary = LinearInterpolate(x0,x1,y0,y1);

exist = true;

end

end

First of all, im not confident that this is the right implementation. The purpose of these lines of code is to, by going through a for-loop, check the existence of this "boundary", and once such a boundary exists, set the variable "exist = true". I need this boolean variable because i need to use it for an if-statement immediately after (if exist == true ... else...).

My problem is that I do not think the codes above are executing the purpose as stated above? I kinda sense something is off but I can't quite tell which part isn't right... Would be really thankful if someone could point to me some ways to correct these codes. Thank you!

Answer 1

[Voss]

If you are using the value of boundary immediately after that for loop, then you are using the value of boundary corresponding to the last "boundary" found within the loop.

It's not clear that this is the problem, but, if, for instance, you want the first boundary, you can break immediately after setting exist = true;, which will exit the loop and go to your if exist ... else ... statements.

If that's not the problem, then please describe what the code does vs what you want it to do.

[Mingze Yin]

Hi, thank you for reply. It should be the second instance u described that I was trying to achieve. Can i check that you meant I should break within the if statement smth like

for ...

if ...

...

exist = true;

break (?)

end

end

[Mingze Yin]

i think there will at most be one such boundary for each j value in the for loop, because based on my observation of the phi values, the function only changes sign at most once every time

[Voss]

Yes, that is where break goes. That will stop looking as soon as one boundary is found.

But if there is actually only one boundary, then something else is likely going on. You mention phi changing signs, so you should maybe consider whether the case that the sign of phi is 0 (e.g., should phi going from positive or negative to 0 be counted as a boundary?).

But I can't say for sure that's the cause of the problem, because you haven't stated what the problematic behavior is. Can you share a concrete example that gives the "wrong" result?

[Mingze Yin]

Actually the thing is I cant directly observe whats wrong with those few lines of codes, as they are only a small part of a larger program. The problem shows in the output of the entire program, and I have trouble pinpointing what is causing the problematic outputs... Therefore I wanted to make sure my implementation for finding the boundary is correct before I try to debug other parts. To talk about the entire program would be an entirely different topic though (and way too specific), so I shall try to figure it out myself first.

Anyway, may I also check with you that, if in the case I do not find a boundary, and still want to exhaust the remaining for-loops, can I write something like

for ...

if ...

...

exist = true;

break

else

continue (?)

end

end

Once again thank you for your patient help!

[Voss]

A continue statement there would be unnecessary and have no effect, since it is at the end of the loop anyway (continue says to move on to the next iteration of the loop).

OK, I understand you are not sure if there is a problem with this part of the code or something else. You may try setting a breakpoint at any part of the code you want to investigate while it's running (so you can check the values of your variables as the code is running). You can read here about how to do that: https://www.mathworks.com/help/matlab/matlab_prog/debugging-process-and-features.html#mw_aac9fccc-eee5-48d5-9c1e-36a598251c93

[Mingze Yin]

Oh thank you so much for this, I actually have been oblivous to this function until now. I tried it and it's already helping me discover some overlooked problems in the program. Thanks again!

[Voss]

You're welcome! That's definitely a useful feature to know about (I use it all the time)