Question

The Perl defined function (and many others) returns “a Boolean value”.

Given Perl doesn’t actually have a Boolean type (and uses values like 1 for true, and 0 or undef for false) does the Perl language specify exactly what is returned for a Boolean values? For example, would defined(undef) return 0 or undef, and is it subject to change?

Answer 1

In almost all cases (i.e. unless there’s a reason to do otherwise), Perl returns one of two statically allocated scalars: &PL_sv_yes (for true) and &PL_sv_no (for false). This is them in detail:

>perl -MDevel::Peek -e"Dump 1==1"
SV = PVNV(0x749be4) at 0x3180b8
  REFCNT = 2147483644
  FLAGS = (PADTMP,IOK,NOK,POK,READONLY,pIOK,pNOK,pPOK)
  IV = 1
  NV = 1
  PV = 0x742dfc "1"\0
  CUR = 1
  LEN = 12

>perl -MDevel::Peek -e"Dump 1==0"
SV = PVNV(0x7e9bcc) at 0x4980a8
  REFCNT = 2147483647
  FLAGS = (PADTMP,IOK,NOK,POK,READONLY,pIOK,pNOK,pPOK)
  IV = 0
  NV = 0
  PV = 0x7e3f0c ""\0
  CUR = 0
  LEN = 12

yes is a triple var (IOK, NOK and POK). It contains a signed integer (IV) equal to 1, a floating point number (NV) equal to 1, and a string (PV) equal to 1.

no is also a triple var (IOK, NOK and POK). It contains a signed integer (IV) equal to 0, a floating point number (NV) equal to 0, and an empty string (PV). This means it stringifies to the empty string, and it numifies to 0. It is neither equivalent to an empty string

>perl -wE"say 0+(1==0);"
0

>perl -wE"say 0+'';"
Argument "" isn't numeric in addition (+) at -e line 1.
0

nor to 0

>perl -wE"say ''.(1==0);"


>perl -wE"say ''.0;"
0

There’s no guarantee that this will always remain the case. And there’s no reason to rely on this. If you need specific values, you can use something like

my $formatted = $result ? '1' : '0';