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From: Jorge on 20 Feb 2010 04:46 On Feb 20, 10:21 am, Asen Bozhilov <asen.bozhi...(a)gmail.com> wrote: > Richard Cornford wrote: > > That is an explanation, but not a full explanation because it does not > > state why +++ must be interpreted as the tokens ++ followed by the token > > +, rather than + followed by ++. The answer to that is in the spec, in > > the last sentence of the first paragraph of section 7, where it says > > "The source text is scanned from left to right, repeatedly taking the > > longest possible sequence of characters as the next input element". So > > given +++, the longest possible sequence of characters that can be a > > token is ++, leaving the last + to be the next token. > > How can you explain: > > var x = 10; > +++x; > > There SyntaxError instead of runtime error. As Richard said, the parser tries to math the longest sequence of characters that makes sense. In this case ++ makes sense, but as soon as the next character found is +, it stops making sense and throws. -- Jorge.
From: Jorge on 20 Feb 2010 04:50 On Feb 20, 10:46 am, Jorge <jo...(a)jorgechamorro.com> wrote: > > As Richard said, the parser tries to math the longest sequence of > characters that makes sense. In this case ++ makes sense, but as soon > as the next character found is +, it stops making sense and throws. s/math/match/ -- Jorge.
From: Asen Bozhilov on 20 Feb 2010 05:36 Jorge wrote: > Asen Bozhilov wrote: > > Richard Cornford wrote: > > > "The source text is scanned from left to right, repeatedly taking the > > > longest possible sequence of characters as the next input element". So > > > given +++, the longest possible sequence of characters that can be a > > > token is ++, leaving the last + to be the next token. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ > > There SyntaxError instead of runtime error. > As Richard said, the parser tries to math the longest sequence of > characters that makes sense. In this case ++ makes sense, but as soon > as the next character found is +, it stops making sense and throws. From which grammar rules `+y` isn't valid `UnaryExpression`? And why in next example there runtime exception instead of SyntaxError? ++ function(){}(); //ReferenceError ++ function(){}; //SyntaxError `++` operator works with ReferenceType "13.2.1 [[Call]]", never return ReferenceType, which is logical because if [[Call]] return ReferenceType will be have reference to Activation Object of execution context which is finish. And AO cannot be marked as garbage collection when execution context exit.
From: Jorge on 20 Feb 2010 07:09 On Feb 20, 11:36 am, Asen Bozhilov <asen.bozhi...(a)gmail.com> wrote: > Jorge wrote: > > Asen Bozhilov wrote: > > > Richard Cornford wrote: > > > > "The source text is scanned from left to right, repeatedly taking the > > > > longest possible sequence of characters as the next input element". So > > > > given +++, the longest possible sequence of characters that can be a > > > > token is ++, leaving the last + to be the next token. > > ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ > > > > There SyntaxError instead of runtime error. > > As Richard said, the parser tries to math the longest sequence of > > characters that makes sense. In this case ++ makes sense, but as soon > > as the next character found is +, it stops making sense and throws. > > From which grammar rules `+y` isn't valid `UnaryExpression`? For the same reason that ++2 throws: +y is a number not a reference. > And why in next example there runtime exception instead of > SyntaxError? > > ++ function(){}(); //ReferenceError > ++ function(){}; //SyntaxError In Safari both throw a referenceError: Prefix ++ operator applied to value that is not a reference. > `++` operator works with ReferenceType "13.2.1 [[Call]]", never return > ReferenceType, which is logical because if [[Call]] return > ReferenceType will be have reference to Activation Object of execution > context which is finish. And AO cannot be marked as garbage collection > when execution context exit. And ? It can return a reference to a function, isn't it? -- Jorge.
From: kangax on 20 Feb 2010 09:58
On 2/20/10 4:21 AM, Asen Bozhilov wrote: > Richard Cornford wrote: > >> That is an explanation, but not a full explanation because it does not >> state why +++ must be interpreted as the tokens ++ followed by the token >> +, rather than + followed by ++. The answer to that is in the spec, in >> the last sentence of the first paragraph of section 7, where it says >> "The source text is scanned from left to right, repeatedly taking the >> longest possible sequence of characters as the next input element". So >> given +++, the longest possible sequence of characters that can be a >> token is ++, leaving the last + to be the next token. > > How can you explain: > > var x = 10; > +++x; > > There SyntaxError instead of runtime error. If this gives SyntaxError, that's a bug in implementation (should be ReferenceError of course, as per PutValue). I know that Mozilla is rather non-conforming in this regard. For example, in Mozilla, CallExpression on the left hand side of AssignmentExpression results in SyntaxError, rather than what should really be a ReferenceError: function f(){} f() = 1; gives "SyntaxError: invalid assignment left-hand side" whereas in WebKit there's an expected "ReferenceError: Left side of assignment is not a reference." -- kangax |