ROUND(1000.2555-999,3)=1.255
the correct result is 1.256,why?

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No, when rounding a value the next number in the figure is considered,
in this case, the 4th decimal (5). If that number is less than 5, then
the result is rounded down, if it's 5 or more, then it's rounded up. So
1.2554 becomes 1.255 but 1.2555 becomes 1.256.

If you want to keep the value without the rounding, use TRUNC instead:

TRUNC(1000.2555-999,3) returns 1.255

Cheers.


liaosheng wrote:
ROUND(1000.2555-999,3)=1.255
the correct result is 1.256,why?

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suggestions with the most votes. To vote for this suggestion, click the "I
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Oops, sorry. misunderstood your post. You are of course correct, it
should return 1.256. No idea why it doesn't.

Cheers


liaosheng wrote:
ROUND(1000.2555-999,3)=1.255
the correct result is 1.256,why?

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The value being rounded is:

1.255499999999980

Biff

"liaosheng" wrote in message
...
ROUND(1000.2555-999,3)=1.255
the correct result is 1.256,why?

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this
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=ROUNDUP(1000.2555-999,3)

mama no teeth

"liaosheng" wrote:

ROUND(1000.2555-999,3)=1.255
the correct result is 1.256,why?

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suggestions with the most votes. To vote for this suggestion, click the "I
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Excel is performing exactly as it should here.

Excel and almost all other computer software does binary math. Most
terminating decimal fractions (including .2555) are non-terminating binary
fractions that can only be approximated (just as 1/3 can only be approximated
as a decimal fraction). The precision of Excel's approximation is defined by
the IEEE 754 standard, where the decimal representation of the best binary
approximation to 1000.2555 is
1000.255499999999983629095368087291717529296875. It displays as 1000.2555
because (as documented in Help) Excel will display no more than 15 digits,
but when you subtract 999, you correctly get a value that is visibly less
than 1.2555, and which therefore rounds down.

Since you know the operations that produced your 1.25549999999998, you know
that ROUND(1000.2555-999,4) will do no violence to your calculations while
cleaning up the result of the original binary approximations. That will
produce a value that will round as you expected.

Jerry

"liaosheng" wrote:

ROUND(1000.2555-999,3)=1.255
the correct result is 1.256,why?

----------------
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suggestions with the most votes. To vote for this suggestion, click the "I
Agree" button in the message pane. If you do not see the button, follow this
link to open the suggestion in the Microsoft Web-based Newsreader and then
click "I Agree" in the message pane.

http://www.microsoft.com/office/comm...et.f unctions