how to round the decimal point?

Hai,
I need to round the decimal number to 2 decimal places. What is the function in matlab, to the above rounding off. Looking for your reply.
BSD

Answers (3)

x = rand * 1000;
y = floor(x * 100) / 100;
See: floor, ceil, round, fix.
A general method to find help is searching in the documentation:
docsearch round

4 Comments

floor() rounds the decimal value to an integer. I should not round it to an integer, I wan't the decimal value to be rounded upto 2 decimal place. For example, 0.1698 must be rounded to 0.17. Looking for your reply.
BSD
In you original question you ask for "rounding off". If you want to round to the nearest: round(0.1698 * 100) / 100 replies 0.17
>> sprintf('%.99g\n',round(0.1698 * 100) / 100)
ans =
0.1700000000000000122124532708767219446599483489990234375
>> num2hex(round(0.1698 * 100) / 100 )
ans =
3fc5c28f5c28f5c3
>> num2hex(round(0.1698 * 100) / 100 - 8*eps(.01))
ans =
3fc5c28f5c28f5c2
If you compare the hex of the two floating point values, you will note that this second value is the first representable value smaller than the original value
>> sprintf('%.99g\n',round(0.1698 * 100) / 100 - 8*eps(0.01))
ans =
0.169999999999999984456877655247808434069156646728515625
and you can see from the extended printout that it is below 0.17 whereas the other value was above 0.17. We have thus experimentally demonstrated that there is no exact representation in binary floating point arithmetic for 0.17 exactly.
@BSD: Does Walter's correct argument concern your work? Or is 0.1700000000000000xyz... accurate enough for your demands?

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Walter Roberson
Walter Roberson on 27 Sep 2011
This is not possible in binary floating point arithmetic, unless the rounded fraction happens to end up being 0.0, 0.25, 0.5, or 0.75
Binary floating point arithmetic cannot exactly represent the fraction 1/100 in any finite amount of storage, for the same reason that decimal arithmetic cannot exactly represent the fraction 1/7 in any finite amount of storage.
Jigar Gada
Jigar Gada on 28 Aug 2012
Thanks a lot.. It was so simple.

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on 27 Sep 2011

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