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What is the biggest difference we can obtain by using these two?

The difference can be seen with Devel::Peek:

#/usr/bin/perl use Devel::Peek; use feature 'say'; my %assign = ( foo => 1234, bar => 5678, ); my %undef = ( foo => 1234, bar => 5678, ); %assign = (); undef %undef; say "%assign (",\%assign,") after assignment:"; Dump(%assign); say "%undef (",\%undef,") after undef:"; Dump(%undef); __END__ %assign (HASH(0x195f890)) after assignment: SV = PVHV(0x193fe60) at 0x195f890 REFCNT = 1 FLAGS = (PADMY,SHAREKEYS) ARRAY = 0x19a1d70 KEYS = 0 FILL = 0 MAX = 7 %undef (HASH(0x195f860)) after undef: SV = PVHV(0x193fe80) at 0x195f860 REFCNT = 1 FLAGS = (PADMY,SHAREKEYS) ARRAY = 0x0 KEYS = 0 FILL = 0 MAX = 7

The difference lies in the ARRAY element of the above output. By assignment of an empty list the container of the key/value pairs is preserved, whereas undef erases it and sets it to a NULL pointer.

It is obviously cheaper to just wipe the container (and its content) instead of iterating through it and re-arranging its internal structure. - edit - see below, what dave_the_m writes.

perl -le'print map{pack c,($-++?1:13)+ord}split//,ESEL'

In reply to Re^3: '%hash = ()' is slower than 'undef %hash' by shmem
in thread '%hash = ()' is slower than 'undef %hash' by rsFalse

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