What does perl stack OPs really do?

llancet has asked for the wisdom of the Perl Monks concerning the following question:

For instance, in this piece of code:

SV* my_sv = newSViv(12345);
XPUSHs(my_sv);
[download]

Does XPUSHs make a copy of my_sv, or it really pushes my_sv into stack?

And in this piece of code:

SPAGAIN;
ENTER;
SAVETMPS;
PUSHMARK(SP);

call_pv( "a_perl_sub_that_return_something", some_flags );

SV* return = POPs;

FREETMPS;
LEAVE;
[download]

It seems if I want to use the poped SV permanently (after FREETMPS and LEAVE), I have to make a copy of return by newSVsv(), otherwise return will lose the value inside it?
Do I need to free the return manually, or it has been done automatically by FREETMPS and LEAVE?

Comment on What does perl stack OPs really do? Select or Download Code

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Re: What does perl stack OPs really do? by ikegami (Patriarch) on Jun 20, 2011 at 18:36 UTC
Does XPUSHs make a copy of my_sv, or it really pushes my_sv into stack? It really pushes it onto the stack. `#define XPUSHs(s) STMT_START { EXTEND(sp,1); (++sp = (s)); } STMT_END` [download] Your code has a memory leak, though. The scalar won't get freed since you didn't mortalise it, meaning you didn't tell Perl the stack owns a reference to it. `SV my_sv = newSViv(12345); XPUSHs(sv_2mortal(my_sv));` [download] Same thing: `SV* my_sv = newSViv(12345); mXPUSHs(my_sv);` [download] It seems if I want to use the poped SV permanently (after FREETMPS and LEAVE), I have to make a copy of return by newSVsv(), I think you can simple increase its REFCNT. If you do this, it's your responsibility to decrement its REFCNT when you're done with it. Do I need to free the return manually, or it has been done automatically by FREETMPS and LEAVE? You just have to call FREETMPS.	[reply] [d/l] [select]
Re^2: What does perl stack OPs really do? by llancet (Friar) on Jun 21, 2011 at 04:34 UTC
I found that perl actually don't return the same scalar. For example: `# perl code { my $static_var = 0; sub return_it { return $static_var } }` [download] If I call this sub from C code several times, each time I will obtain returned SVs that have different address. The SVs are observed instantly after pop: `SV* tmp = POPs; Perl_sv_dump(my_perl,tmp);` [download] Why? Does perl already made a copy of return values?	[reply] [d/l] [select]
Re^3: What does perl stack OPs really do? by Anonymous Monk on Jun 23, 2011 at 09:48 UTC
I found that perl actually don't return the same scalar... Doubtful. Can you copy/paste the output? Or better still, a patch , something along the lines of `$ diff -ruN tmp.empty tmp.full diff -ruN tmp.empty/1 tmp.full/1 --- tmp.empty/1 1969-12-31 16:00:00.000000000 -0800 +++ tmp.full/1 2011-06-23 02:46:59.875000000 -0700 @@ -0,0 +1 @@ +ECHO is on. diff -ruN tmp.empty/2 tmp.full/2 --- tmp.empty/2 1969-12-31 16:00:00.000000000 -0800 +++ tmp.full/2 2011-06-23 02:47:01.312500000 -0700 @@ -0,0 +1 @@ +ECHO is on.` [download] so that we may compile code try the same code as you :)	[reply] [d/l]
Re^4: What does perl stack OPs really do? by llancet (Friar) on Jun 25, 2011 at 02:20 UTC
Re^4: What does perl stack OPs really do? by llancet (Friar) on Jun 25, 2011 at 01:52 UTC
Re: What does perl stack OPs really do? by Anonymous Monk on Jun 20, 2011 at 09:52 UTC
Does XPUSHs make a copy I don't think so. perlxstut and http://perldoc.perl.org/perlguts.html#XSUBs-and-the-Argument-Stack lead me to believe it only copies a reference. or it has been done automatically by FREETMPS and LEAVE? http://perldoc.perl.org/perlcall.html#Passing-Parameters says: Think of these macros as working a bit like { and } in Perl to limit the scope of local (read mortal) variables.	[reply]