CODEWITHSUNDEEP
javascriptd3.jsqueue
user3164565
user3164565

Reputation: 67

queue.js not asynchronously?

I want to run several tasks in parallel and use queue.js of mbostock.I run the following code with using Sleep(),so expecting the other lighter tasks to be finished before the heavy tasks while all of the tasks executed at the same time.But following code result in executing tasks sequentially

2 1 test.html:128
1000 2 test.html:134
4 3 test.html:128
3000 4 test.html:134
6 5 test.html:128

while I was expecting something like(execute every tasks at same time but lighter tasks finish earlier): 

2 1 test.html:128 
4 3 test.html:128
6 5 test.html:128    
1000 2 test.html:134
3000 4 test.html:134

What am I doing wrong? code:

function Sleep(ms) {
    var d1 = new Date().getTime();
    var d2 = new Date().getTime();
    while( d2 < (d1 + ms) ) {
        d2 = new Date().getTime();
    }
    return;
}

var tmp_list = [];
var normal = function(src){
    tmp_list.push(src);
    console.log(src,tmp_list.length);
}

var sleepy = function(src){
    Sleep(5000);
    tmp_list.push(src);
    console.log(src,tmp_list.length)
};

queue().defer(normal,2)
        .defer(sleepy,1000)
        .defer(normal,4)
        .defer(sleepy,3000)
        .defer(normal,6)
        .awaitAll(function(){});

Upvotes: 1

Views: 71

Answers (1)

musically_ut
musically_ut

Reputation: 34288

Javascript's model for concurrency is a bit different from other languages.

In Javascript, everything runs synchronously except input/output. This input/output usually takes the shape of AJAX calls. You can also explicitly yield control using the setTimeout function.

In your case, you are creating a busy loop while in Sleep function:

function Sleep(ms) {
    var d1 = new Date().getTime();
    var d2 = new Date().getTime();
    while( d2 < (d1 + ms) ) {
        d2 = new Date().getTime();
    }
    return;
}

This will keep the CPU busy humming along and not yield control until it is done. However, if you truly want to wait for the task to be over, you have to make use of the callback which the queue library provides to the function being called and use setTimeout:

 /* cb :: function (err, result) is added by queue library to the arg list */
function sleepy = function (src, cb) {
    setTimeout(function () {
        tmp_list.push(src);
        console.log(src, tmp_list.length);
        cb(); // no errors
    }, 5000);
}

Also, I suspect that your function after .awaitAll is not getting called either (since you are not using queue's callbacks). Try putting a console.log in there to verify this.

Upvotes: 1

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