将参数传递给setTimeout回调函数

Question

I have some JS code as below;

var x = self.someAJAXResponseJSON; // x has some object value here
setTimeout(function(x){
    console.log("setTimeout ... : " + x); // But x is undefined here
}, 1000);

So I want to pass "x" to the setTimeout callback function. But I am getting "x" as undefined inside the setTimeout.

What am I doing wrong ?

UPDATED

Any idea of fix for similar issue using DOJO JS

setTimeout(dojo.hitch(this, function(){
    this.executeSomeFunction(x); // what shud be this
    console.log("setTimeout ... : " + x); // But x is undefined here
}), 1000);

Answer 1

Alternatively you can do it without creating a closure.

function myFunction(str1, str2) {
  alert(str1); //hello
  alert(str2); //world
}

window.setTimeout(myFunction, 10, 'hello', 'world');

But note it doesn't work on IE < 10 according to MDN .

Answer 2

When setTimeout invokes the callback, it doesn't pass any arguments (by default); that is, the argument x is undefined when the callback is invoked.

If you remove the parameter x , x in the function body won't refer to the undefined parameter but instead to the variable you defined outside the call to setTimeout() .

var x = "hello";
setTimeout(function () { //note: no 'x' parameter
    console.log("setTimeout ... : " + x);
}, 1000);

Alternatively, if it must be a parameter, you can pass it as an argument to setTimeout (do yourself a favor and name it differently, though):

var x = "hello";
setTimeout(function (y) {
    console.log("setTimeout ... : " + y);
}, 1000, x);

Answer 3

Ran into this myself and looked at the Node docs, arguments to be passed into the function come in as a 3rd(or more) parameter to the setTimeout call so...

myfunc = function(x){console.log(x)};
x = "test";
setTimeout(myfunc,100,x);

Worked for me.

Answer 4

In your code, console.log(x) refers to the x parameter of the callback function.

Just omit it from function signature, and you'll be fine:

setTimeout(function(){
  console.log("setTimeout ... : " + x); // now x is the global x
}, 1000);

Answer 5

It is because the function is called without passing it any argument: so x is undefined.

You should wrap it in a closure if you are willing to call it with different parameters for x :

var x = self.someAJAXResponseJSON; // x has some object value here
setTimeout((function(y){
    return(function() {
        console.log("setTimeout ... : " + y);
    })
})(x), 1000);

Answer 6

setTimeout method is designed to take function or code snippet as its first argument but in you case you have taken an anonymous function with parameter x and which is called without taking any argument so it shows is undefined because there is no concrete function defined that takes x. What you can do is you can first define the function you want to call and then just call it in setTimeout method. See following code snippet:

var x = self.someAJAXResponseJSON;
function mylog(x){
  console.log("setTimeout ... : " + x);
}
setTimeOut(mylog(x),1000);