1 //! When specifying SSR rule, you generally want to map one *kind* of thing to
2 //! the same kind of thing: path to path, expression to expression, type to
5 //! The problem is, while this *kind* is generally obvious to the human, the ide
6 //! needs to determine it somehow. We do this in a stupid way -- by pasting SSR
7 //! rule into different contexts and checking what works.
9 use syntax::{ast, AstNode, SyntaxNode};
11 pub(crate) fn ty(s: &str) -> Result<SyntaxNode, ()> {
12 let template = "type T = {};";
13 let input = template.replace("{}", s);
14 let parse = syntax::SourceFile::parse(&input);
15 if !parse.errors().is_empty() {
18 let node = parse.tree().syntax().descendants().find_map(ast::Type::cast).ok_or(())?;
19 if node.to_string() != s {
22 Ok(node.syntax().clone())
25 pub(crate) fn item(s: &str) -> Result<SyntaxNode, ()> {
27 let input = template.replace("{}", s);
28 let parse = syntax::SourceFile::parse(&input);
29 if !parse.errors().is_empty() {
32 let node = parse.tree().syntax().descendants().find_map(ast::Item::cast).ok_or(())?;
33 if node.to_string() != s {
36 Ok(node.syntax().clone())
39 pub(crate) fn expr(s: &str) -> Result<SyntaxNode, ()> {
40 let template = "const _: () = {};";
41 let input = template.replace("{}", s);
42 let parse = syntax::SourceFile::parse(&input);
43 if !parse.errors().is_empty() {
46 let node = parse.tree().syntax().descendants().find_map(ast::Expr::cast).ok_or(())?;
47 if node.to_string() != s {
50 Ok(node.syntax().clone())