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Fix std.HashMap.remove returning incorrect KV
Now returns a copy of the removed kv instead of a pointer to the removed kv. The removed kv gets overwritten when shifting the hash map after the removal, so returning a pointer to it will have another kv's values in it after the return. This bug had some nasty downstream effects in things like BufSet and BufMap where delete would free a still in-use KV and leave the actually removed KV un-free'd.
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@ -175,7 +175,7 @@ pub fn HashMap(comptime K: type, comptime V: type, comptime hash: fn (key: K) u3
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return hm.get(key) != null;
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}
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pub fn remove(hm: *Self, key: K) ?*KV {
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pub fn remove(hm: *Self, key: K) ?KV {
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if (hm.entries.len == 0) return null;
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hm.incrementModificationCount();
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const start_index = hm.keyToIndex(key);
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@ -189,13 +189,14 @@ pub fn HashMap(comptime K: type, comptime V: type, comptime hash: fn (key: K) u3
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if (!eql(entry.kv.key, key)) continue;
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const removed_kv = entry.kv;
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while (roll_over < hm.entries.len) : (roll_over += 1) {
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const next_index = (start_index + roll_over + 1) % hm.entries.len;
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const next_entry = &hm.entries[next_index];
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if (!next_entry.used or next_entry.distance_from_start_index == 0) {
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entry.used = false;
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hm.size -= 1;
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return &entry.kv;
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return removed_kv;
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}
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entry.* = next_entry.*;
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entry.distance_from_start_index -= 1;
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@ -371,7 +372,10 @@ test "basic hash map usage" {
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testing.expect(map.contains(2));
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testing.expect(map.get(2).?.value == 22);
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_ = map.remove(2);
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const rmv1 = map.remove(2);
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testing.expect(rmv1.?.key == 2);
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testing.expect(rmv1.?.value == 22);
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testing.expect(map.remove(2) == null);
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testing.expect(map.get(2) == null);
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}
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