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On Mon, Nov 27, 2017 at 3:44 PM, Michael Orlitzky <mjo@g.o> wrote: |
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> On 11/27/2017 03:37 PM, Arve Barsnes wrote: |
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> > |
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> > Sounds kind of weird? If he has keyworded the game package, shouldn't it |
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> > just never install that version if it depends on an unstable package? |
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> |
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> That's right, but if there are two available ~arch versions, one of |
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> which has all stable dependencies and (the newer) one of which has all |
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> ~arch dependencies, then portage will try to install the newer one and |
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> tell you to keyword a million things -- even though it could install the |
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> first one with less hassle. |
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> |
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> |
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In theory the resolver could support such a scheme, no? |
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Currently it prefers "best_visible" where best == "highest version" |
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But other predicates might be possible like "tightest_visible" |
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Note that currently best_visible is fairly straightforward to implement: |
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visible = find_all_visible() |
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best_visible = max(visible) # simple ver_cmp |
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tightest_visible: |
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visible = find_all_visible() |
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tightest_visible = min(visible, key=cpv.tightness) |
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# A weird predicate where 'tightness' implies dependencies are already |
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visible? I can't think of a better term. |
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def tightness(self): |
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return len([atom for atom in self.dependencies if atom.is_visible()]) |
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So "tighter" packages have dependencies that are visible / installed. |
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"Loose" packages have unmet dependencies or dependencies that are not |
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currently visible. |
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I have not messed with the portage resolver in some time; in theory given |
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perfect code you just plum this in once.... |
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but I'm skeptical its that easy (and I suspect best_visible is in fact |
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'presumed' in many places. |
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But Zac would probably know best. |
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-A |
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> For example, if you're on a system with no ruby packages, then it's the |
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> difference between having to keyword 10 packages for rails-x.y.z versus |
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> 200 packages for rails-x.y.(z+1). I'd rather have the slightly older |
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> version that requires less configuration. |
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> |
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> |
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> |
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