🐛 No-op left transforms where the inserted objects are identical
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dawidreedsy
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Oct 3, 2024
| @@ -669,7 +669,7 @@ json.transformComponent = function(dest, c, otherC, type) { | |||
| } else if (otherC.oi !== void 0) { | |||
| if (c.oi !== void 0 && c.p[common] === otherC.p[common]) { | |||
| // left wins if we try to insert at the same place | |||
| if (type === 'left') { | |||
| if (type === 'left' && !deepEqual(otherC.oi, c.oi)) { | |||
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Good prompt: I added a test case and realised that ops with no od are handled by different code to ops with an od 💀
However, apart from the quirks of this codebase, I don't think it matters if they have different od: they should still no-op. Ultimately, transform(a, b) is saying: "make a compatible with a document where b has already been applied." So if we've already applied b, its oi is the current value. Applying a — with the same oi — would result in no change, regardless of what its od is. Therefore, it should be a no-op.
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test/json0.coffee
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| it 'An attempt to re-add a key with an identical string becomes a no-op', -> | ||
| assert.deepEqual [], type.transform [{p:['k'], oi:'x'}], [{p:['k'], oi:'x'}], 'left' | ||
| assert.deepEqual [], type.transform [{p:['k'], oi:'x'}], [{p:['k'], oi:'x'}], 'right' | ||
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| it 'An attempt to re-replace a key with an identical string becomes a no-op', -> | ||
| assert.deepEqual [], type.transform [{p:['k'], oi:'x', od: 'a'}], [{p:['k'], oi:'x', od: 'a'}], 'left' | ||
| assert.deepEqual [], type.transform [{p:['k'], oi:'x', od: 'a'}], [{p:['k'], oi:'x', od: 'a'}], 'right' | ||
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| it 'An attempt to re-add a key with an identical object becomes a no-op', -> | ||
| assert.deepEqual [], type.transform [{p:['k'], oi:{}}], [{p:['k'], oi:{}}], 'left' | ||
| assert.deepEqual [], type.transform [{p:['k'], oi:{}}], [{p:['k'], oi:{}}], 'right' | ||
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| it 'An attempt to re-replace a key with an identical object becomes a no-op', -> | ||
| assert.deepEqual [], type.transform [{p:['k'], oi:{}, od: 'a'}], [{p:['k'], oi:{}, od: 'a'}], 'left' | ||
| assert.deepEqual [], type.transform [{p:['k'], oi:{}, od: 'a'}], [{p:['k'], oi:{}, od: 'a'}], 'right' | ||
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Let's add a test case for different od
At the moment if you transform two identical `oi` ops by one another, you get a no-op in the `right` case, but the `left` case results in a "silly" op whose `od` value is identical to its `oi` value. `left` vs `right` should simply determine which op wins in a tie. Since the two ops are identical, the `right` scenario and `left` scenario should also be identical. This change updates the transform behaviour to check this case, and results in a no-op regardless of `left` or `right` if the two ops are performing identical `oi`.
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At the moment if you transform two identical
oiops by one another, you get a no-op in therightcase, but theleftcase results in a "silly" op whoseodvalue is identical to itsoivalue.leftvsrightshould simply determine which op wins in a tie. Since the two ops are identical, therightscenario andleftscenario should also be identical.This change updates the transform behaviour to check this case, and results in a no-op regardless of
leftorrightif the two ops are performing identicaloi.