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Postings from January 2011
2011/1/26 John Peacock <john.peacock@havurah-software.org>: > On 01/26/2011 12:06 PM, Brad Baxter wrote: >> >> Perl *is* storing '1 x' as a number. Very interesting. > > Not exactly: > > $ perl -MDevel::Peek -E '$b = "1 x"; say Dump($b); 0 + $b; say Dump($b); say > $b."yz"' > SV = PV(0x14aceb68) at 0x14aee1d0 > REFCNT = 1 > FLAGS = (POK,pPOK) > PV = 0x14ae80f0 "1 x"\0 > CUR = 3 > LEN = 8 > > SV = PVNV(0x14ad0290) at 0x14aee1d0 > REFCNT = 1 > FLAGS = (IOK,NOK,POK,pIOK,pNOK,pPOK) > IV = 1 > NV = 1 > PV = 0x14ae80f0 "1 x"\0 > CUR = 3 > LEN = 8 > > 1 xyz > > > The first type we dump $b is as a pure "string" where the second time, $b is > now both a string and a number (actually an integer and a floating point). > As you can see, if you perform an explicit string operation line catenate, > the PV value is used, not the IV or NV... You probably miss -w $ perl -we'print 0 + "1 x"' Argument "1 x" isn't numeric in addition (+) at -e line 1. 1 $ perl -e'print 0 + "1 x"' 1 -- Reini Urban http://phpwiki.org/ http://murbreak.at/Thread Previous | Thread Next