comment on

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
[download]

Update: BTW, same catch applies to %hashes.

In reply to Re: Why does foo() evaluate in array context in "${\foo()}"? by repellent
in thread Why does foo() evaluate in array context in "${\foo()}"? by ELISHEVA

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