# On multiple logical test failing in if statement

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Tailman on 8 Jul 2018
Edited: Tailman on 9 Jul 2018
Hello, I am trying to run an if statement with multiple conditions as below,
if abs(a-a_t)>0.0001 && abs(b-b_t)>0.00001 && abs(c-c_t)>0.0001 && abs(d-d_t)>0.00001
x=0;
y=0;
end
Even when all the conditions are satisfied, the values of x and y are not updated. But, it works if we have just two conditions (I've tried different combinations of two variable values which work, but using all conditions in separate brackets doesn't work). MATLAB documentation says that multiple logical operators could be used, but it doesn't seem to be working in my case. Is there a limit to the number of logical operators that could be combined? Or, am I missing something?

Stephen Cobeldick on 8 Jul 2018
"Is there a limit to the number of logical operators that could be combined?"
No.
"Or, am I missing something?"
Probably. How are you checking the output? Please show us the exact input values (all eight of them). Please upload your code by clicking the paperclip button.
Tailman on 8 Jul 2018
Hello Stephen,
This is puzzling, not to mention embarrassing too. I tried to run the code again and found that it is running. I am unsure what was the problem last time (it is the exact same code and inputs as back then). Thanks again for your reply, though!

dpb on 8 Jul 2018
Edited: dpb on 8 Jul 2018
Apparently the issue was likely one of the cases wasn't actually T or some similar misstep, but illustrates the difficulties inherent in writing code with sequentially-named variables. Use arrays instead.
If had defined a variable data and commensurate test value, then the following could be written much more succinctly and much easier to both code and debug--
data=[a b c ...]; % the data array (*)
test=[a_t b_t c_t ...]; % associated test values
lims=[0.0001 0.00001, 0.0001, ...]; % and criteria
if all(abs(data-test))<lims % see if all ok
....
(*) And, of course, this is not intended to suggest one should write the specific line concatenating all these variables to create the one; instead use the vector features of Matlab when creating the variables initially, either by reading one or more files or the result of other calculations, whatever.
Don't create the problem in the beginning to have to try to fix later...

Image Analyst on 9 Jul 2018
Well x, x_temp, y, y_temp, z, and z_temp each do what they do independent of each other (i.e. x and x_temp don't depend on y and z, or a, b, c, or d). So each loop for x, y, and z is independent and the order doesn't matter as far as I can see. There is no break or continue so each must get done in turn.
Likewise, the big if statement doesn't depend on x, x_temp, y, y_temp, z, or z_temp so it's independent and will get tested no matter what.
I don't know what "most sensitive" means (maybe first to converge?) but if you're talking about short circuiting and not testing b, c, and d if abs(a-a_temp)>0.0001, then that's not true. It will test for b, c, and d because you have && in there not ||. Anyway, even if it was || and you did short circuit and not test the others, you'd only save like a half a nanosecond or something - not enough to worry about.
Tailman on 9 Jul 2018
Thank you for pointing this out. It should be an "or" condition. I am also adding breakpoints to see if there's something else I'm missing.
Tailman on 9 Jul 2018
Sorry, I deleted the previous comment instead of editing it (Gist- I used && because by the time the first value converges, remaining three too should have converged, so, when abs(a-a_t)<0.0001, everything else too should be lesser than 0.0001).
Yes, I meant that the first value would converge finally. Thus, if the first condition were to be satisfied, all the remainder should be satisfied by nature of the problem. But, still, I think that using "or" condition would be the better option, in case something unexpected comes up.
Also, I thought that if the first condition were not met, then && would skip the remainder, as the logical statement would anyway result in a 0, whereas the "or" would go on to check the remainder as we could have a result of 1. Would this be the other way around?

R2017b

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