Closed Bug 1670495 Opened 4 years ago Closed 4 years ago

In strict mode code, |x = y| needs to throw a ReferenceError if |x| isn't declared, even if evaluating |y| would create |x|

Categories

(Core :: JavaScript Engine, defect, P3)

Product:
Core
Component:
JavaScript Engine
Type:
defect

Tracking

()

Status:
RESOLVED DUPLICATE of bug 605515

People

(Reporter: Waldo, Unassigned)

References

(Blocks 1 open bug)

Details

The latest test262 says that

"use strict"; undeclared = (this.undeclared = 5);

should throw a ReferenceError, but we don't right now. Filing a bug to get on the radar, and for citing in jstests.list...

Blocks: test262
Severity: -- → S4
Priority: -- → P3
Status: NEW → RESOLVED
Closed: 4 years ago
Resolution: --- → DUPLICATE
You need to log in before you can comment on or make changes to this bug.