stevedlawrence commented on code in PR #1065:
URL: https://github.com/apache/daffodil/pull/1065#discussion_r1496417269
##########
daffodil-core/src/main/scala/org/apache/daffodil/core/compiler/Compiler.scala:
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@@ -325,7 +325,7 @@ class Compiler private (
optRootName: Option[String] = None,
optRootNamespace: Option[String] = None,
): ProcessorFactory = {
- val source = URISchemaSource(file.toURI)
+ val source = URISchemaSource((file.getPath, file.toURI))
Review Comment:
I guess maybe it adds an extra level in indirection without much gain? For
example, if `URISchemaSource` looked like this:
```scala
class URISchemaSource(val schemaOption)
```
Then to create it you need to create a separate object, e.g.
```scala
val schemaOption = new SchemaOption(..., ...)
val schemaSource = new URISchemaSource(schemaOption)
```
And then either everything needs to be changed to do
`schemaSource.schemaOption.uri` for example, or `URISchemaSource` needs to
update its members to reach into `schemaOption` for he uri and diagnostics.
Really, the only reason why we need `SchemaOption` is because scallop
doesn't have a way to access to the original CLI argument after it's been
converted. So you either need the a tuple (or a new object that's basically a
tuple) to keep the original string around. Nothing else in the code really
needs this pair of diagnostic + uri separately/ And that's really all the
URISchemaSource it's, it's a tuple of the URI and how to show that URI in
diagnostics plus some helper stuff.
Which makes me wonder if instead of creating a `URI` in the
`fileResourceURIConverter`, we should just create a `URISchemaSource` directly.
I imagine everything that uses the `schema` option as a `URI` just creates a
`URISchemaSource` anyways. And if it needs the `URI`, it can just access the
`uri` member of the `URISchemaSource`?
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