Fix #4498: Go to definition in lifted expressions #4516
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Consider the following snippet:
```
val lst = new A :: Nil
```
This is rewritten as:
```
val lst = {
val x$0 = new A
Nil.::(x$0)
}
```
During lifting, we need to assign positions to the trees. We were
assigning the position of `new A` in the original code to `x$0` in
`Nil.::(x$0)`, and the original position of `new A` to `val x$0 = new A`
with a zero-extent.
Still considering this snippet, this means that when trying to go to the
definition on `new A`, the IDE would think that the user tried to go to
the definition of `x$0`, because the IDE would look for the matching
tree by position.
This commit changes this so that the lifted definition
(`val x$0 = new A`) gets the position of the original expression, and
the reference gets a synthetic, zero-extent position. With this change,
the IDE correctly finds the right tree when selecting `new A` in the
original code.
Fixes scala#4498
|
@smarter I think you're the positions expert, can you take a look at this patch, please? |
Interesting, I didn't think of this. We should check what happens when going to definition on accessors created by the Inliner because I used a similar pattern there: 87ffb88 |
smarter
approved these changes
May 11, 2018
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Consider the following snippet:
This is rewritten as:
During lifting, we need to assign positions to the trees. We were
assigning the position of
new Ain the original code tox$0inNil.::(x$0), and the original position ofnew Atoval x$0 = new Awith a zero-extent.
Still considering this snippet, this means that when trying to go to the
definition on
new A, the IDE would think that the user tried to go tothe definition of
x$0, because the IDE would look for the matchingtree by position.
This commit changes this so that the lifted definition
(
val x$0 = new A) gets the position of the original expression, andthe reference gets a synthetic, zero-extent position. With this change,
the IDE correctly finds the right tree when selecting
new Ain theoriginal code.
Fixes #4498