Zakas' new Quiz
in [JavaScript]
|
Prev: FAQ Topic - Why does framename.print() not print the correct frame in IE? (2010-02-19)
Next: My Library TaskSpeed tests updated--MultipleIE has the IE6 problem,not My Library
From: Thomas 'PointedEars' Lahn on 20 Feb 2010 00:43 Garrett Smith wrote: > Richard Cornford wrote: >> Asen Bozhilov wrote: >>> Garrett Smith wrote: >>>> Example #4 >>> I was give answer TypeError on this question. >> For a system that does not implement a - substr - method for strings? > > DMD Script, maybe? The DMDScript source code in D indicates that it implements String.prototype.substr(): <http://www.digitalmars.com/dscript/> PointedEars -- realism: HTML 4.01 Strict evangelism: XHTML 1.0 Strict madness: XHTML 1.1 as application/xhtml+xml -- Bjoern Hoehrmann
From: kangax on 20 Feb 2010 02:09 On 2/19/10 8:10 PM, Richard Cornford wrote: > Garrett Smith wrote: >> Yet another JS Quiz: >> http://www.nczonline.net/blog/2010/02/16/my-javascript-quiz/ >> >> I have a couple of comments on the explanations provided: >> http://www.nczonline.net/blog/2010/02/18/my-javascript-quiz-answers/ >> >> Example #1: >> I believe the explanation is that the postfix is matched, as in: >> >> TOKENS: num1, ++, +, num2; > > 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. This also explains why something like `x+++++x` can not be parsed as if it was `x++ + ++x`. First, `++` that follows `x` is consumed. Next longest token is again `++`, leaving only `+` and `x`. The whole sequence becomes: `x`, `++`, `++`, `+`, and `x` (`x++ ++ +x`), which is obviously not a valid program. As soon as we prevent 3rd and 4th (or even just 4th) `+` from becoming one token, expression parses as expected: `x+++ ++x`. [...] -- kangax
From: Asen Bozhilov on 20 Feb 2010 03:57 Richard Cornford wrote: > Asen Bozhilov wrote: > > I was give answer TypeError on this question. > > For a system that does not implement a - substr - method for strings? Exactly. `String.prototype.substr' isn't part from any editions of ECMA-262 standard. On this questions `TypeError' is possible answer. Another possible answer is: "I don't know." and again is correct answer. | 2 Conformance | [...] | In particular, a conforming implementation of | ECMAScript is permitted to provide properties not described in this specification, | and values for those properties, | for objects that are described in this specification. The standard permit methods like `substr', but these methods are implementation depended, because concrete implementation can implement in own way. I don't think methods like `substr' should be part from quizzes, because confuse readers, especially when author of quiz isn't specified concrete environment.
From: Asen Bozhilov on 20 Feb 2010 04:21 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.
From: Jorge on 20 Feb 2010 04:43
On Feb 19, 10:38 pm, Garrett Smith <dhtmlkitc...(a)gmail.com> wrote: > > Example #2: > The explanation provided mentions scope, and that is totally wrong. The > scope of the anonymous function expression wrapped in the setTimeout is > (anonymous)[[Scope]] --> doIt [[Scope]] -- global [[Scope]]. Yep, every function carries its context with it, that's what closures are about, and as soon as a reference to a function survives longer than the context in which it was created (as when saved in the setTimeout queue), a closure is born. > The answer as to why `this.x` === 5 is explained by the fact that the > execution context for the nonstandard `setTimeout` method is global context. No. Not at all. It's explained by the fact that "this" is === window. It has nothing to do with the execution context. It could have been very well any other context !== the global one, and still "this" would have been "window". E.g. if you wrapped it into yet another function. -- Jorge. |