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POHow *exactly* does the RHS of PowerShell's -f operator work?
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copied!<p><a href="https://stackoverflow.com/questions/1827862/what-determines-whether-the-powershell-pipeline-will-unroll-a-collection">Last time I got confused</a> by the way <a href="http://en.wikipedia.org/wiki/Windows_PowerShell" rel="nofollow noreferrer">PowerShell</a> eagerly unrolls collections, Keith summarized its heuristic like so:</p> <blockquote> <p>Putting the results (an array) within a grouping expression (or subexpression e.g. $()) makes it eligible again for unrolling.</p> </blockquote> <p>I've taken that advice to heart, but still find myself unable to explain a few esoterica. In particular, the Format operator doesn't seem to play by the rules.</p> <pre><code>$lhs = "{0} {1}" filter Identity { $_ } filter Square { ($_, $_) } filter Wrap { (,$_) } filter SquareAndWrap { (,($_, $_)) } $rhs = "a" | Square # 1. all succeed $lhs -f $rhs $lhs -f ($rhs) $lhs -f $($rhs) $lhs -f @($rhs) $rhs = "a" | Square | Wrap # 2. all succeed $lhs -f $rhs $lhs -f ($rhs) $lhs -f $($rhs) $lhs -f @($rhs) $rhs = "a" | SquareAndWrap # 3. all succeed $lhs -f $rhs $lhs -f ($rhs) $lhs -f $($rhs) $lhs -f @($rhs) $rhs = "a", "b" | SquareAndWrap # 4. all succeed by coercing the inner array to the string "System.Object[]" $lhs -f $rhs $lhs -f ($rhs) $lhs -f $($rhs) $lhs -f @($rhs) "a" | Square | % { # 5. all fail $lhs -f $_ $lhs -f ($_) $lhs -f @($_) $lhs -f $($_) } "a", "b" | Square | % { # 6. all fail $lhs -f $_ $lhs -f ($_) $lhs -f @($_) $lhs -f $($_) } "a" | Square | Wrap | % { # 7. all fail $lhs -f $_ $lhs -f ($_) $lhs -f @($_) $lhs -f $($_) } "a", "b" | Square | Wrap | % { # 8. all fail $lhs -f $_ $lhs -f ($_) $lhs -f @($_) $lhs -f $($_) } "a" | SquareAndWrap | % { # 9. only @() and $() succeed $lhs -f $_ $lhs -f ($_) $lhs -f @($_) $lhs -f $($_) } "a", "b" | SquareAndWrap | % { # 10. only $() succeeds $lhs -f $_ $lhs -f ($_) $lhs -f @($_) $lhs -f $($_) } </code></pre> <p>Applying the same patterns we saw in the previous question, it's clear why cases like #1 and #5 behave different: the pipeline operator signals the script engine to unroll another level, while the assignment operator does not. Put another way, everything that lies between two |'s is treated as a grouped expression, just as if it were inside ()'s.</p> <pre><code># all of these output 2 ("a" | Square).count # explicitly grouped ("a" | Square | measure).count # grouped by pipes ("a" | Square | Identity).count # pipe + () ("a" | Square | Identity | measure).count # pipe + pipe </code></pre> <p>For the same reason, case #7 is no improvement over #5. Any attempt to add an extra <strong>Wrap</strong> will be immediately subverted by the extra pipe. Ditto #8 vs #6. A little frustrating, but I'm totally on board up to this point. </p> <p>Remaining questions:</p> <ul> <li>Why doesn't case #3 suffer the same fate as #4? <strong>$rhs</strong> should hold the nested array <strong>(,("a", "a"))</strong> but its outer level is getting unrolled...somewhere...</li> <li>What's going on with the various grouping operators in #9-10? Why do they behave so erratically, and why are they needed at all?</li> <li>Why don't the failures in case #10 degrade gracefully like #4 does?</li> </ul>

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