2021-12-27 10:44:45 -06:00
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//! When specifying SSR rule, you generally want to map one *kind* of thing to
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//! the same kind of thing: path to path, expression to expression, type to
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//! type.
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//!
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//! The problem is, while this *kind* is generally obvious to the human, the ide
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//! needs to determine it somehow. We do this in a stupid way -- by pasting SSR
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//! rule into different contexts and checking what works.
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use syntax::{ast, AstNode, SyntaxNode};
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pub(crate) fn ty(s: &str) -> Result<SyntaxNode, ()> {
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let template = "type T = {};";
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let input = template.replace("{}", s);
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let parse = syntax::SourceFile::parse(&input);
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if !parse.errors().is_empty() {
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return Err(());
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}
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let node = parse.tree().syntax().descendants().find_map(ast::Type::cast).ok_or(())?;
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2021-12-27 12:05:26 -06:00
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if node.to_string() != s {
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return Err(());
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}
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2021-12-27 10:44:45 -06:00
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Ok(node.syntax().clone())
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}
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2021-12-27 11:33:33 -06:00
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pub(crate) fn item(s: &str) -> Result<SyntaxNode, ()> {
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let template = "{}";
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let input = template.replace("{}", s);
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let parse = syntax::SourceFile::parse(&input);
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if !parse.errors().is_empty() {
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return Err(());
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}
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let node = parse.tree().syntax().descendants().find_map(ast::Item::cast).ok_or(())?;
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2021-12-27 12:05:26 -06:00
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if node.to_string() != s {
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return Err(());
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}
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2021-12-27 11:33:33 -06:00
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Ok(node.syntax().clone())
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}
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2021-12-27 12:13:55 -06:00
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pub(crate) fn expr(s: &str) -> Result<SyntaxNode, ()> {
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let template = "const _: () = {};";
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let input = template.replace("{}", s);
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let parse = syntax::SourceFile::parse(&input);
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if !parse.errors().is_empty() {
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return Err(());
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}
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let node = parse.tree().syntax().descendants().find_map(ast::Expr::cast).ok_or(())?;
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if node.to_string() != s {
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return Err(());
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}
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Ok(node.syntax().clone())
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}
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