http://www.perlmonks.org/?node_id=933450 use strict;use Devel::Peek;my $a;Dump($a);$a=4;Dump($a);$a eq "13";Dump($a);$a++;Dump($a);print $a;Dump($a);####C:\Users\ian>perl test.plSV = NULL(0x0) at 0x1d9ec1c REFCNT = 1 FLAGS = (PADMY)SV = IV(0x1d9ec18) at 0x1d9ec1c REFCNT = 1 FLAGS = (PADMY,IOK,pIOK) IV = 4SV = PVIV(0x1d91c34) at 0x1d9ec1c REFCNT = 1 FLAGS = (PADMY,IOK,POK,pIOK,pPOK) IV = 4 PV = 0x1dbdf84 "4"\0 CUR = 1 LEN = 4SV = PVIV(0x1d91c34) at 0x1d9ec1c REFCNT = 1 FLAGS = (PADMY,IOK,pIOK) IV = 5 PV = 0x1dbdf84 "4"\0 CUR = 1 LEN = 4SV = PVIV(0x1d91c34) at 0x1d9ec1c REFCNT = 1 FLAGS = (PADMY,IOK,POK,pIOK,pPOK) IV = 5 PV = 0x1dbdf84 "5"\0 CUR = 1 LEN = 45####SV = NULL(0x0) at 0x1d9ec1c REFCNT = 1 FLAGS = (PADMY)####SV = IV(0x1d9ec18) at 0x1d9ec1c REFCNT = 1 FLAGS = (PADMY,IOK,pIOK) IV = 4####SV = PVIV(0x1d91c34) at 0x1d9ec1c REFCNT = 1 FLAGS = (PADMY,IOK,POK,pIOK,pPOK) IV = 4 PV = 0x1dbdf84 "4"\0 CUR = 1 LEN = 4####SV = PVIV(0x1d91c34) at 0x1d9ec1c REFCNT = 1 FLAGS = (PADMY,IOK,pIOK) IV = 5 PV = 0x1dbdf84 "4"\0 CUR = 1 LEN = 4####SV = PVIV(0x1d91c34) at 0x1d9ec1c REFCNT = 1 FLAGS = (PADMY,IOK,POK,pIOK,pPOK) IV = 5 PV = 0x1dbdf84 "5"\0 CUR = 1 LEN = 4
presents some of the details of how perl stores and keeps track of variables and their values. It is a bit more complex than the simple model presended in and the commonly used terms "variable" and "value". There are several data structures used and linked to each other in the case of most variables. Depending on the history of the variable, there will be various bits of memory allocated and with more or less current values.
You can follow the evolution of $a at each step in your program by adding a few more Dump calls:
Which produces:
The first Dump produces:
At this point, the variable is declared but no value has been assigned. There is a root data structure allocated: SV, but no memory allocated to store a value: the value is undef and a NULL pointer value in the SV structure is sufficient to record this.
After assigning an integer value to the variable, Dump produces:
Here an IV structure has been allocated and the assigned value, 4, has been stored in it. The pointer in the SV structure now points to the IV structure: it is not a NULL pointer any more. Flags are set to indicate that the variable now contains a valid integer value (IOK and pIOK).
Next you do a string comparison, for which perl needs to convert the variable value to a string. After this Dump shows:
The string value could be created, used then discarded, but instead perl updates the stored value, keeping the string value in case it might be used again. The integer value is still correct, so the variable now has two values: an integer value and a string value. Two data structures are used: the original SV and a PVIV structure which contains the integer value and a pointer to the string value. And the flags are updated to indicate that there are valid integer and string values.
Your next operation is to increment $a. This is an operation on its integer value. In this case, perl is frugal and does only enough work to record the result. The integer value is changed but the string value is not changed. Rather, the flags indicating a valid string value are turned off. After this, Dump produces:
Note that perl does not bother to change the PVIV structure back to an IV structure. The latter is smaller but the saving of memory would be tiny - not enough to justify the work. So, the PVIV structure remains, and it still points to the memory allocated to hold the string value, which still has the old, but no longer correct string value of the variable. It's OK that the string value is incorrect because the POK and pPOK flags are off, so the value will not be used. This wastes some memory, but saves some CPU.
Then you print the value of $a. Again, a string value is needed. While the PVIV structure has a pointer to a string value, the POK and pPOK flags in the SV are off, so this value isn't used. Instead, perl again converts the current integer value to a string and prints that correct value. And, again, perl saves this value in case it might be used again. So, after printing Dump produces:
Again there are both integer and string values, and they are current and consistent. The IOK, POK, pIOK and pPOK flags are all set so that in subsequent operations these value might be used.
Note also that the new string value is in the same bit of memory that the original string value was in. No additional memory was needed, so no new memory was allocated. The original memory was re-used. A lot of work has gone into minimizing expensive operations for the sake of performance.
You might try extending your program, assigning a long string value to $a and then dumping it again. If you did, you would see that the same SV and PVIV structures were in use but the IOK and pIOK flags were off and the integer value in the PVIV remained unchanged and the long string was contained in newly allocated memory, as it wouldn't fit in the block originally allocated to hold the previous single character value.