# can anyone suggest me a command that talks about how matlab performed an arithmetic operation (its steps to give us the final result) ?

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diadalina on 25 Sep 2018
Commented: Bruno Luong on 30 Sep 2018
such a given operation (1/2+1/3)/456*789
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Steven Lord on 30 Sep 2018
People have commented about potential ways to do what you've asked, but no one has yet asked why you want to know this information?

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### Accepted Answer

Stephen23 on 25 Sep 2018
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Stephen23 on 30 Sep 2018
"if i want to use the the debugging tools.for this operation how can i do this"
I don't know if it is possible to do exactly what you want. I wrote that the closest solution might be to use the debugging tools (in particular the step in command). But how useful that would be depends on what commands in your code are functions and which ones are compiled/inbuilt. But there is nothing stopping you from giving it a try:

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### More Answers (2)

Bruno Luong on 26 Sep 2018
Edited: Bruno Luong on 26 Sep 2018
Here is an idea, replace the number of your expression by functions (that return a scalar) and print out the order
% (1/2+0/3)/456*789
(f(1)/f(2)+f(0)/f(3))/f(456)*f(789)
function x = f(x)
disp(x)
end
When run it you'll get
>> printorder
1
2
0
3
456
789
ans =
0.8651
Insert the operator(s) (there is only one right way to do it) you'll get the polish-reverse notation of the expression
1; 2; /; 0; 3; /; +; 456; /; 789; *
Meaning the expected specified evaluation order we told you.
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Bruno Luong on 30 Sep 2018
From the order of the evaluation of the operands one can infer the order of the operations. Actually I have double checked the logic by hacking the operations as well to display something when it's invoked. The result just match well all the cases I tested with what I have expected and the documentation.

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Bruno Luong on 26 Sep 2018
Edited: Bruno Luong on 26 Sep 2018
f(3)^f(4) would have to be done first
That's the case, the power is the first to be performed (among operators), but it's not saying the operands f(3) and f(4) are evaluated first.
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Bruno Luong on 26 Sep 2018
sorry my reply should be in the comment, not a new answer

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