Powershell [int][Math]::Floor($a/$b) versus [Math]::Floor([int]$a / [int]$b)
<# | |
In which I prove that that the terser [int][Math]::Floor($a/$b) gives the | |
same results as the technet recommended [Math]::Floor([int]$a / [int]$b) | |
Said technet recomendation is available at: | |
http://technet.microsoft.com/en-us/library/ee176879.aspx | |
One extra cast is probably slower too. | |
#> | |
1.. 1000 | ForEach-Object { | |
Foreach ($divisor in 2,3,5,7) { | |
$a = [int][Math]::Floor($_ / $divisor) | |
$b = [Math]::Floor([int]$_ / [int]$divisor) | |
if ($a -ne $b) { | |
"$($a) <> $($b) : $($a)/$($b)" | |
} | |
} | |
} |
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