Question

66

Getting scope of function caller

rated 0 times [ 70] [ 4] / answers: 1 / hits: 17324 / 13 Years ago, fri, september 16, 2011, 12:00:00

I have a function that breaks somewhere in Line 1433 of ExtJS.

var createDelayed = function(h, o, scope){

console.log(arguments); //logs undefined all round. 

    return function(){

        var args = Array.prototype.slice.call(arguments, 0);

        setTimeout(function(){

            h.apply(scope, args);

        }, o.delay || 10);

    };

};

Is there any way to see what line a function is executed from, from within itself?

(since it's a third party lib, and I cant really do

var me =this;

and log me)

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masonm

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answered 13 Years ago dannyc · Accepted Answer

There is arguments.callee.caller, which refers to the function that called the function in which you access that property. arguments.callee is the function itself.

There is no way to get the scope of the original function without passing it. In the following example, you cannot determine the this value inside foo (apart from knowing there is nothing special happening with this here):

function foo() {

    bar();

}



function bar() {

    console.log(arguments.callee);        // bar function

    console.log(arguments.callee.caller); // foo function

}



foo();

Documentation

To get the line number things becomes trickier, but you can throw an error and look at the stack trace: http://jsfiddle.net/pimvdb/6C47r/.

function foo() {

    bar();

}



function bar() {

    try { throw new Error; }

    catch(e) {

        console.log(e.stack);

    }

}



foo();

For the fiddle, it logs something similar to the following in Chrome, where the end of the line says the line number and character position:

Error

    at bar (http://fiddle.jshell.net/pimvdb/6C47r/show/:23:17)

    at foo (http://fiddle.jshell.net/pimvdb/6C47r/show/:19:5)

    at http://fiddle.jshell.net/pimvdb/6C47r/show/:29:1