What does it mean if change the order of code bugs occur?

Question

I change places function so that the caller is before called, an error occurs.

What can say this character? Maybe somewhere in the distance of a kilometer bracket is not closed?

UPD: For example, its code work correctly if I place first string at end:

SCRIPT5022: Pass a function that returns the value of the dependentObservable knockout-2.0.0.debug.js, line 1054 character 9

osagoViewModel.fields.yearsBoxes = new field("Years", yearsBoxesFunc, null, osagoViewModel);


function yearsBox() {
    this.year = new field("Years", function () { return ["1 year", "2 years", "3 years", "4 years", "5 years", "6 years", "7 years", "8 years", "9 years", "10 years"]; }, null, osagoViewModel);
}


var yearsBoxesFunc = function () {
    var yearsBoxCount = osagoViewModel.fields.driversCount.selectedValue();

    var retArrFunc = function (count) {
        var arr = [];
        for (var i = 0; i < count; i++) {
            arr.push(new yearsBox());
        }
        return arr;
    };


    switch (yearsBoxCount) {
        case "many":
            return retArrFunc(0);
        case "1":
            return retArrFunc(1);
        case "2":
            return retArrFunc(2);
        case "3":
            return retArrFunc(3);
        case "4":
            return retArrFunc(4);
        case "5":
            return retArrFunc(5);
    }
}

If bracket in a distance of a km missing is, to long code you have — mplungjan, May 11 '12 at 13:01
that was unintentionally funnier than I expected, seeing as it _appears_ the question is about the order in which things are parsed... My comment was about the content of the question itself! ;-) — Alnitak, May 11 '12 at 13:01

score 1 · Answer 1 · answered May 11 '12 at 12:54

it depends how you declare the function. there is a difference between if you use var or not. have a look on this sample:

function FunctionDefinitionOrder() {
    assert(isRed(), "Named function definition doesn't matter");        
    assert(isGreen === undefined, "However, it's not the case with anonymous functions");

    function isRed() { return true; }
    var isGreen = function () { return true; };

    assert(isGreen(), "Anonymous functions are known only after the definition");
}

score 1 · Answer 2 · answered May 11 '12 at 12:55

1

You should revert your changes and be on the latest stable state. And then start making minor changes whatever you want to do. This will take a lot of time to identify any syntax error in the code.

answered May 11 '12 at 12:55

Tahir

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score 1 · Accepted Answer · edited May 23 '17 at 12:11

1

Open your console and let it halt on errors. In the stacktrace you will see where you called the knockout.js-function without a function as an argument.

To read about parsing of functions and their availability in the current scope, read What is the difference between a function expression vs declaration in JavaScript? and/or var functionName = function() {} vs function functionName() {}.

edited May 23 '17 at 12:11

Community

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answered May 11 '12 at 13:45

Bergi

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What does it mean if change the order of code bugs occur?

3 Answers3