Hi,

On my 64-bit Windows 7 OS, I have:
1) perl-5.32.0 built using a 32-bit gcc-8.3.0 compiler;
2) perl-5.32.0 built using a 64-bit gcc-8.3.0 compiler.

On both of those perls, the ivtype is the 8-byte "long long".

On the 32-bit build I get:
C:\>perl -MDevel::Peek -e "Dump 1 << 62;" SV = IV(0x1dc8d64) at 0x1dc8d64 REFCNT = 1 FLAGS = (PADTMP,IOK,READONLY,PROTECT,pIOK) IV = 4611686018427387904
Note that the first line contains two occurrences of the same hex value - 0x1dc8d64 in this instance. Every time I run that command on that 32-bit build of perl, the two hex values are always equivalent - though, of course, they're not always 0x1dc8d64.

When I run the same command on the 64-bit build, I get:
C:\>perl -MDevel::Peek -le "Dump 1 << 62;" SV = IV(0x33d430) at 0x33d440 REFCNT = 1 FLAGS = (PADTMP,IOK,READONLY,PROTECT,pIOK) IV = 4611686018427387904
For this build of perl, it is always the case that the first hex value is less than the second value by 0x10.

How is this differing behaviour between the 32-bit build and the 64-bit build explained ?

IIUC, on the 64-bit build, the amount of memory being allocated for the IV is 0x10 (16) bytes more than is necessary ... and I wonder what is responsible for that extravagance.
Is it perl ? ... or the OS ? ... or the compiler ? ...

Cheers,
Rob

In reply to Understanding Devel::Peek output by syphilis

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