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Should PUSHSTACKi() set the mark? #13331
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From @nwc10As best I can tell, PUSHSTACKi() is completely undocumented. It pushes a new If I understand it correctly, the mark stack marks places on the perl stack. #define PUSHMARK(p) \ and that last assignment is using pointer arithmetic to calculate the offset. Now, when PUSHSTACKi() does its stuff, it calls SWITCHSTACK, which does this: #define SWITCHSTACK(f,t) \ ie, use a different array. So at this point, the index represented by the This strikes me as a bit of gotcha. So should we change SWITCHSTACK() to Nicholas Clark |
From @cpansproutOn Fri Oct 04 05:14:19 2013, nicholas wrote:
Anything popping the caller’s mark is already buggy, so having a bad Unless you meant pushing a new markstack, but that may be overkill. -- Father Chrysostomos |
The RT System itself - Status changed from 'new' to 'open' |
From @iabynOn Fri, Oct 04, 2013 at 06:16:27AM -0700, Father Chrysostomos via RT wrote:
+1 Also, when dumping the stack with -Ds and -Dsv, the si_markoff field The only thing that makes me slightly twitchy is the bare -- |
From @cpansproutOn Fri Oct 04 15:36:19 2013, davem wrote:
I think that means I can mark this as rejected. -- Father Chrysostomos |
@cpansprout - Status changed from 'open' to 'rejected' |
Migrated from rt.perl.org#120100 (status was 'rejected')
Searchable as RT120100$
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