Same thing would happen if you did my $STRING = ${ +shift }; if you had passed a reference. If you don't want a copy, don't make a copy! (Also, don't use a 20 year old version of the language!) This was already all covered in detail. Including how what you claim happens hasn't been true for 10 years.

In reply to Re^4: Passing argument by reference (for a scalar) by ikegami
in thread Passing argument by reference (for a scalar) by jmClifford

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