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# Test parameter subscripting.
%prep
s='Twinkle, twinkle, little *, [how] I [wonder] what? You are!'
a=('1' ']' '?' '\2' '\]' '\?' '\\3' '\\]' '\\?' '\\\4' '\\\]' '\\\?')
typeset -g -A A
A=($a)
%test
x=','
print $s[(i)winkle] $s[(I)winkle]
print ${s[(i)You are]} $#s
print ${s[(r)$x,(R)$x]}
0:Scalar pattern subscripts without wildcards
>2 11
>53 60
>, twinkle, little *,
x='*'
print $s[(i)*] $s[(i)\*] $s[(i)$x*] $s[(i)${(q)x}*] $s[(I)$x\*]
print $s[(r)?,(R)\?] $s[(r)\?,(R)?]
print $s[(r)\*,(R)*]
print $s[(r)\],(R)\[]
0:Scalar pattern subscripts with wildcards
>1 26 1 26 26
>Twinkle, twinkle, little *, [how] I [wonder] what? ? You are!
>*, [how] I [wonder] what? You are!
>] I [
# $s[(R)x] actually is $s[0], but zsh treats 0 as 1 for subscripting.
print $s[(i)x] : $s[(I)x]
print $s[(r)x] : $s[(R)x]
0:Scalar pattern subscripts that do not match
>61 : 0
>: T
print -R $s[$s[(i)\[]] $s[(i)$s[(r)\*]] $s[(i)${(q)s[(r)\]]}]
0:Scalar subscripting using a pattern subscript to get the index
>[ 1 33
print -R $a[(r)?] $a[(R)?]
print $a[(n:2:i)?] $a[(n:2:I)?]
print $a[(i)\?] $a[(I)\?]
print $a[(i)*] $a[(i)\*]
0:Array pattern subscripts
>1 ?
>2 2
>3 3
>1 13
# It'd be nice to do some of the following with (r), but we run into
# limitations of the ztst script parsing of backslashes in the output.
print -R $a[(i)\\\\?] $a[(i)\\\\\?]
print -R $a[(i)\\\\\\\\?] $a[(i)\\\\\\\\\?]
print -R ${a[(i)\\\\\\\\?]} ${a[(i)\\\\\\\\\?]}
print -R "$a[(i)\\\\\\\\?] $a[(i)\\\\\\\\\?]"
print -R $a[(i)\]] $a[(i)\\\\\]] $a[(i)\\\\\\\\\]] $a[(i)\\\\\\\\\\\\\]]
print -R $a[(i)${(q)a[5]}] $a[(i)${(q)a[8]}] $a[(i)${(q)a[11]}]
print -R $a[(i)${a[3]}] $a[(i)${a[6]}] $a[(i)${a[9]}] $a[(i)${a[12]}]
0:Array pattern subscripts with multiple backslashes
>4 6
>7 9
>7 9
>7 9
>2 5 8 11
>5 8 11
>1 3 4 6
print -R $A[1] $A[?] $A[\\\\3] $A[\\\]]
print -R $A[$a[11]]
print -R $A[${(q)a[5]}]
0:Associative array lookup (direct subscripting)
>] \2 \\] \?
>\\\?
>\\\?
# The (o) is necessary here for predictable output ordering
print -R $A[(I)\?] ${(o)A[(I)?]}
print -R $A[(i)\\\\\\\\3]
print -R $A[(I)\\\\\\\\\?] ${(o)A[(I)\\\\\\\\?]}
0:Associative array lookup (pattern subscripting)
>? 1 ?
>\\3
>\\? \\3 \\?
print -R $A[(R)\?] : ${(o)A[(R)?]}
print -R $A[(R)\\\\\?] ${(o)A[(R)\\\\?]} ${(o)A[(R)\\\\\?]}
print -R ${(o)A[(R)\\\\\\\\\]]}
0:Associative array lookup (reverse subscripting)
>: ]
>\? \2 \? \?
>\\]
x='*'
A[$x]=xstar
A[${(q)x}]=qxstar
print -R ${(k)A[(r)xstar]} $A[$x]
print -R ${(k)A[(r)qxstar]} $A[${(q)x}]
# A[*] is interpreted specially, assignment to it fails silently (oops)
A[*]=star
A[\*]=backstar
print -R ${(k)A[(r)star]} $A[$x]
print -R ${(k)A[(r)backstar]} $A[\*]
0:Associative array assignment
>* xstar
>\* qxstar
>xstar
>\* backstar
o='['
c=']'
A[\]]=cbrack
A[\[]=obrack
A[\\\[]=backobrack
A[\\\]]=backcbrack
print -R $A[$o] $A[$c] $A[\[] $A[\]] $A[\\\[] $A[\\\]]
print -R $A[(i)\[] $A[(i)\]] $A[(i)\\\\\[] $A[(i)\\\\\]]
0:Associative array keys with open and close brackets
>obrack cbrack obrack cbrack backobrack backcbrack
>[ ] \[ \]
print -R $A[$o] $A[$s[(r)\[]]
print -R $A[(r)$c] $A[(r)$s[(r)\]]]
print -R $A[$A[(i)\\\\\]]]
0:Associative array lookup using a pattern subscript to get the key
>obrack obrack
>] ]
>backcbrack
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