The way I see it,
${ EXPR1 } forces scalar context
and then performs a scalar dereference on
EXPR1.
The backslash
\( EXPR2 ) forces list context on
EXPR2 and returns a list of references.
By combining the two, you have
\foo() returning a list of one reference to scalar
10. Then,
${ (\10) } returns the scalar dereferenced
10.
FYI, there's a catch to using the
\( ) syntax for
@arrays:
> perl -de ''
DB<1> x @array = (1 .. 3)
0 1
1 2
2 3
DB<2> x \(@array, (@array))
0 ARRAY(0x844cad8)
0 1
1 2
2 3
1 SCALAR(0x844e420)
-> 1
2 SCALAR(0x844ca3c)
-> 2
3 SCALAR(0x844cb80)
-> 3
DB<3> x \(@array)
0 SCALAR(0x844e420)
-> 1
1 SCALAR(0x844ca3c)
-> 2
2 SCALAR(0x844cb80)
-> 3
DB<4> x \(@array, ())
0 ARRAY(0x844cad8)
0 1
1 2
2 3
Update: BTW, same catch applies to
%hashes.
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