[Parser] Parse empty interpolations

Previously, the parser had an odd asymmetry which caused the same function to accept foo(), but reject “\()”. This change fixes the issue.

Already tested by test/Parse/try.swift, which uses this construct in one of its throwing interpolation tests.

Author: beccadax

lib/Parse/Parser.cpp

@@ -917,6 +917,10 @@ ParserStatus
 Parser::parseList(tok RightK, SourceLoc LeftLoc, SourceLoc &RightLoc,
                   bool AllowSepAfterLast, Diag<> ErrorDiag, SyntaxKind Kind,
                   llvm::function_ref<ParserStatus()> callback) {
+  auto TokIsStringInterpolationEOF = [&]() -> bool {
+    return Tok.is(tok::eof) && Tok.getText() == ")" && RightK == tok::r_paren;
+  };
+  
   llvm::Optional<SyntaxParsingContext> ListContext;
   ListContext.emplace(SyntaxContext, Kind);
   if (Kind == SyntaxKind::Unknown)
@@ -929,6 +933,10 @@ Parser::parseList(tok RightK, SourceLoc LeftLoc, SourceLoc &RightLoc,
     RightLoc = consumeToken(RightK);
     return makeParserSuccess();
   }
+  if (TokIsStringInterpolationEOF()) {
+    RightLoc = Tok.getLoc();
+    return makeParserSuccess();
+  }
 
   ParserStatus Status;
   while (true) {
@@ -948,7 +956,7 @@ Parser::parseList(tok RightK, SourceLoc LeftLoc, SourceLoc &RightLoc,
     // If the lexer stopped with an EOF token whose spelling is ")", then this
     // is actually the tuple that is a string literal interpolation context.
     // Just accept the ")" and build the tuple as we usually do.
-    if (Tok.is(tok::eof) && Tok.getText() == ")" && RightK == tok::r_paren) {
+    if (TokIsStringInterpolationEOF()) {
       RightLoc = Tok.getLoc();
       return Status;
     }