Removing decimal places after second number

Question

I have some raw data exported from my ERP. Most of the data has two decimals, but there are a few lines contain numbers with three or more decimal places... When summing up all data with powershell I have a 0,01€ discrepancy between my ERP PDF report and sum of exported data. I tracked the discrepancy to a single number - 2.205. Powershell rounds this number to 2.2 not 2.21 as expected. If I try any other number - 2.215, 2.225 they all get rounded to 2.22, 2.23..

Why is 2.205 not treated the same???

Answer 1

.NET's default mid-point rounding strategy is ToEven, meaning that .5 values are rounded to the closest even integer.

Important: The following example use [decimal] numbers - as implied by number-literal suffix d - not [double]s , so as to avoid problems that stem from [double] values, which internally use a binary representation that typically does not have an exact decimal equivalent, so that a number that may appear to have a 5 in the relevant decimal place may actually be just above or below that exact decimal value,^[1] which means that mid-point rounding may not apply - see Rounding and precision.

In the case at hand, rounding -2.205 to 2 decimal places means that 0 is the closest even integer, and [Math]::Round(-2.205d, 2) therefore yields -2.20.^[2]

It sounds like you want AwayFromZero as the mid-point rounding strategy, which rounds to the next higher integer in absolute terms:

PS> [Math]::Round(-2.205d, 2, 'AwayFromZero')
-2.21

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^{[1] To see if a given fractional value has an exact decimal representation, you can use something like the following:
$num = [double] -2.205; '{0:F28}' -f $num
Unless the input number is echoed with trailing zeros only, the given value has no exact decimal representation; [double] value -2.205 does not, but 0.5 does, for instance.}

^{[2] Note that [Math]::Round(2.225d, 2) therefore yields 2.22, not 2.23, as you state in your question.}