## how to round the decimal point?

### bsd (view profile)

on 27 Sep 2011

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

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### Jan Simon (view profile)

Answer by Jan Simon

### Jan Simon (view profile)

on 27 Sep 2011

```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
```

Jan Simon

### Jan Simon (view profile)

on 27 Sep 2011

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

Walter Roberson

### Walter Roberson (view profile)

on 27 Sep 2011

>> 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.

Jan Simon

### Jan Simon (view profile)

on 28 Sep 2011

@BSD: Does Walter's correct argument concern your work? Or is 0.1700000000000000xyz... accurate enough for your demands?

### Walter Roberson (view profile)

Answer by Walter Roberson

### Walter Roberson (view profile)

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 (view profile)

on 28 Aug 2012

Thanks a lot.. It was so simple.

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