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[iter] Simplify operator!= of iterator filters 

Both to save ops, and also because lambdas don't implement operator!=,
so this was failing in range-based for loop if a lambda was passed to
hb_map() or hb_filter().  Just check end-condition assuming that we
are comparing to .end() or iterators that are otherwise derived from
current iterator.  Ie. don't compare things that are expected to be
in common.
This commit is contained in:
Behdad Esfahbod 2019-05-15 15:14:26 -07:00
parent d3e1d5044f
commit ebf47a95f2
1 changed files with 5 additions and 3 deletions
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8
src/hb-iter.hh
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@ -378,7 +378,7 @@ struct hb_map_iter_t :
void __rewind__ (unsigned n) { it -= n; }
hb_map_iter_t __end__ () const { return hb_map_iter_t (it.end (), f); }
bool operator != (const hb_map_iter_t& o) const
{ return it != o.it || f != o.f; }
{ return it != o.it; }
private:
Iter it;
@ -441,7 +441,7 @@ struct hb_filter_iter_t :
void __prev__ () { do --it; while (it && !hb_has (p.get (), hb_get (f.get (), *it))); }
hb_filter_iter_t __end__ () const { return hb_filter_iter_t (it.end (), p, f); }
bool operator != (const hb_filter_iter_t& o) const
{ return it != o.it || p != o.p || f != o.f; }
{ return it != o.it; }
private:
Iter it;
@ -553,6 +553,8 @@ struct hb_zip_iter_t :
void __prev__ () { --a; --b; }
void __rewind__ (unsigned n) { a -= n; b -= n; }
hb_zip_iter_t __end__ () const { return hb_zip_iter_t (a.end (), b.end ()); }
/* Note, we should stop if ANY of the iters reaches end. As such two compare
* unequal if both items are unequal, NOT if either is unequal. */
bool operator != (const hb_zip_iter_t& o) const
{ return a != o.a && b != o.b; }
@ -621,7 +623,7 @@ struct hb_counter_iter_t :
void __rewind__ (unsigned n) { v -= n * step; }
hb_counter_iter_t __end__ () const { return hb_counter_iter_t (end_, end_, step); }
bool operator != (const hb_counter_iter_t& o) const
{ return v != o.v || end_ != o.end_ || step != o.step; }
{ return v != o.v; }
private:
static inline T end_for (T start, T end_, S step)