What is the easiest way to do integer division in PowerShell? So that the arguments are integers and the result is also an integer.
In C++ it would be e.g. 115/10.
1 Answer
For true integer division, use System.Math.Truncate and cast the result to [int]:
[int] [Math]::Truncate(115 / 10) # -> 11
Alternatively, use System.Math.DivRem, which directly returns an [int] (or [long]):
[Math]::DivRem(115, 10, [ref] $null) # -> 11
In PowerShell (Core) 7+, additional, integer-type specific overloads are available that return tuples, so you may alternatively use:
# PowerShell 7+ only.
[Math]::DivRem(115, 10)[0] # -> 11
PowerShell widens numerical types on demand, and switches to [double] for division with integer operands (of any integer type, such as [int]) that would yield a remainder.
Casting to the result of a division to [int] (e.g. [int] (115 / 10)) is not the same as using [Math]::Truncate(), as that performs half-to-even rounding - see this answer for details.
115/10would be adoublein powershell so now you need to tell us which type of rounding do you prefer to convert that double into an integer. It might be as easy as115 / 10 -as [int]or you might need to use[Math]::Roundwith theMidpointRoundingoverload.