cue/parser: use more robust logic for detecting the end of an interpolation

The current logic for parsing a string interpolation assumes that
an open-parenthesis implies an expression inside the
interpolation, but that's not necessarily the case
and the current logic panics when that's not the case
(although that panic is hidden by the `defer` logic in the
parser).

Fix the logic by checking the end of the string segment: it
can only be an open-paren when the string starts a new
interpolation expression.

Signed-off-by: Roger Peppe <[email protected]>
Change-Id: Ib9cc874a78aec455b9597d8387cd5e0f20885f82
Reviewed-on: https://cue.gerrithub.io/c/cue-lang/cue/+/1223785
TryBot-Result: CUEcueckoo <[email protected]>
Reviewed-by: Daniel Martí <[email protected]>
Unity-Result: CUE porcuepine <[email protected]>

Author: rogpeppe

cue/parser/parser.go

@@ -1595,7 +1595,9 @@ func (p *parser) parseInterpolation() (expr ast.Expr) {
 	last := &ast.BasicLit{ValuePos: pos, Kind: token.STRING, Value: lit}
 	exprs := []ast.Expr{last}
 
-	for p.tok == token.LPAREN {
+	// Note: we can only tell if the string returned by ResumeInterpolation
+	// starts a new interpolated expression by whether it ends in a parenthesis.
+	for strings.HasSuffix(last.Value, "(") {
 		c.pos = 1
 		p.expect(token.LPAREN)
 		cc.closeExpr(p, last)

cue/parser/parser_test.go

@@ -572,9 +572,7 @@ cannot import package as definition identifier`,
 		{
 			desc: "paren after interpolation",
 			in:   `"\(1)"()`,
-			// Note: this actually masks an out-of-bounds slice index panic
-			// in scanner.Scanner.popInterpolation.
-			out: "\nexpected operand, found ')' (and 1 more errors)",
+			out:  `"\(1)"()`,
 		},
 		{
 			desc: "file comments",