Perhaps because "${X()}${X()}" does ${X()} . ${X()} before any stringification makes a copy, so both aliases to the hidden $x are incremented again before their values are copied to the new string. While the other case turns into ( ${X()} . ' ' ) . ${X()} so the first alias to the hidden $x gets copied into a string with a trailing space before $x gets incremented again.

Lots of surprises to discover when you do such things.

- tye        

In reply to Re: Order of evaluation/interpolation of references (alias vs copy) by tye
in thread Order of evaluation/interpolation of references by Anonymous Monk

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