Vista - function to delete a function

How would a function be written that given a name of a function, would
delete that function?

I tried with no success

function rif { ri function:$Args[0] }

- Larry

This seems to work:

function rif { param($f) ri function:$f }

"Larry__Weiss" wrote:

Consider these two versions of function testArgs0_1

function testArgs0_1 { '$Args[0]' + ' ' + $Args[0] + ' ' + '$Args[1]' +
' ' +
$Args[1]}
testArgs0_1 xxx yyy
produces
$Args[0] xxx $Args[1] yyy

function testArgs0_1 { '$Args[0]' + ' ' + "$Args[0]" + ' ' + '$Args[1]'
+ ' '
+ "$Args[1]"}
testArgs0_1 xxx yyy
produces
$Args[0] xxx yyy[0] $Args[1] xxx yyy[1]

Why does putting double quotes around $Args[0] and $Args[1] make that
difference?

- Larry

function testArgs0_1 { $Args0 = $Args[0]; $Args1 = $Args[1]; '$Args[0]'
+ ' ' + "$Args0" + ' ' + '$Args[1]' + ' ' + "$Args1"}
testArgs0_1 xxx yyy
produces
$Args[0] xxx $Args[1] yyy

and works as I expected it to as it avoids a dynamic string expansion
involving $Args[0]

The dynamic string expander seems to stop at the [ character instead of
parsing
the entire $Args[0] token for replacement.

- Larry



Larry__Weiss wrote:+ ' ' + $Args[1]}'$Args[1]' + ' ' + "$Args[1]"}difference?

(I'm so sorry about the HTML versions of my postings...here is a version
of my last post now in plaintext)


function testArgs0_1 { $Args0 = $Args[0]; $Args1 = $Args[1]; '$Args[0]'
+ ' ' + "$Args0" + ' ' + '$Args[1]' + ' ' + "$Args1"}
testArgs0_1 xxx yyy
produces
$Args[0] xxx $Args[1] yyy

and works as I expected it to as it avoids a dynamic string expansion
involving $Args[0]

The dynamic string expander seems to stop at the [ character instead of
parsing the entire $Args[0] token for replacement.

Consider these two versions of function testArgs0_1

function testArgs0_1 { '$Args[0]' + ' ' + $Args[0] + ' ' + '$Args[1]'
+ ' ' + $Args[1]}
testArgs0_1 xxx yyy
produces
$Args[0] xxx $Args[1] yyy

function testArgs0_1 { '$Args[0]' + ' ' + "$Args[0]" + ' ' + '$Args[1]'
+ ' ' + "$Args[1]"}
testArgs0_1 xxx yyy
produces
$Args[0] xxx yyy[0] $Args[1] xxx yyy[1]

Why does putting double quotes around $Args[0] and $Args[1] make that
difference?

- Larry

I figured it out. You have to wrap it with a $() block as in

function testArgs0_1 { '$Args[0]' + ' ' + "$($Args[0])" + ' ' +
'$Args[1]' + ' ' + "$($Args[1])"}
testArgs0_1 xxx yyy
produces
$Args[0] xxx $Args[1] yyy

- Larry



Larry__Weiss wrote: