Rounding decimals matlab manipulates

I need to write, in 1 code line, given a number (pos, neg, int, float), how do I round it without preset functions like fix, round, etc.?

9 Comments

How would you do it in multiple lines of code?
hint 1: watch out for negative
hint 2: log10 to determine the number of decimal places it already has
hint 3: watch out for 0
hint 4: watch out for integers that are exact powers of 10: for example, log10(100) = 2 exactly, log10(99.999) = slightly less than 2.
hint 5: x * (1-eps) for positive x is always less than x
yuval ohayon's "Answer" moved here:
Well thank for the hints Accualy in one code line zerp loops. Smone can help?
@Yuval,
a) Use Comment on this question rather than starting an answer
b) proofread what you read. The above is incomprehensible.
Jan
Jan on 1 Aug 2017
Edited: Jan on 1 Aug 2017
@yuval ohayon: "without the preset function like fix,round" is not clear enough. Is "rem" and "mod" allowed or not? If so, simply use "rem(..., 1)".
Is floor() permitted?
Im not familier with rem function,floor isn permitted. For my understanding it must be with mod func,but how i find if the numbet is upper .5 or down .5
Someones help please
In general people on MATLAB Answers don't post answers to homework questions or questions that sound like homework. If you show what you've done to try to solve the problem on your own and ask a specific question about where you're having difficulty you may receive some more specific suggestions.
Of course, if this isn't a homework problem, just use the round function or the other rounding functions included in MATLAB.
mod() is also a built in function like round(), fix(), int32(), ceil(), floor(), etc.

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Answers (1)

Try it:
x = 13.3
y = mod(x, 1)
z = 13.6
y = mod(x, 1)
Now you should see how to use the output to decide, if the rest of the division by 1 is lower or higher than 0.5.

3 Comments

Well ,how i make condision with no loop x=input number Y=x-xmod(x,1)
Help
x - mod(x,1) + mod(x,1)>0.5
int64() rounds also. And you can use cast() to preserve the original class.

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Asked:

on 31 Jul 2017

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on 2 Aug 2017

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