can anyone please help me to figure out what this code javaScript code means
(function(){
for(var i=0;i<5;i++){
setTimeout(console.log(i),1000);
}
})();
3 Answers
You have run into very common closure issue. To fix this you can have for example self invoked function. Also you should pass function handler to setTimeout instead of invoking console.log:
for(var i=0;i<5;i++){
(function( i ) {
setTimeout( function( ) {
console.log(i);
},1000);
})( i );
}
If you want to print to console numbers from 0 to 4 in 1000ms interval you should use setInterval function:
var intervalHandler = null
, i = 0;
intervalHandler = setInterval( function() {
console.log( i );
if( i === 4 ) {
clearInterval( intervalHandler );
return;
}
i++;
}, 1000 );
7 Comments
Constantine
are you sure this will print the numbers to the console with a interval of 1000ms ?
antyrat
This will not print them w/ interval of 1000ms, this will print them all after 1000ms. If you want to print numbers w/ interval you will need another functionality for this and probably use
setInterval method or recursive setTimeout.Constantine
your comment is pretty close can you please elaborate it as i can understand it
antyrat
@Constantine don't forget to accept answers if they was helpful and welcome to SO ;) Cheers.
Constantine
i am sorry i dont have enough rank to rate up your comments
|
Your code basically calls a function which is gonna log
0
1
2
3
4
A single time, why ? Because setTimeout is actually running setTimeout(undefined, 1000) 5 times, according to the return value of console.log, it will not evaluate a function, so the instruction is just lost.
Though from what I understand of what the code tries to say, to make the code work well, you could delegate the iteration control to a self calling function delayed with setSimeout
(function self(times, delay) {
self.config = self.config || {
times: times,
delay: delay,
iteration: 0
};
++self.config.iteration;
if (self.config.iteration <= self.config.times) {
console.log(self.config.iteration);
setTimeout(self, self.config.delay);
}
})(5, 1000);
5 Comments
Constantine
thanks 'aduch' your answer works as expected but i want to know why is not my code running or what does it means to the js engine
Constantine
your answer really defines the reason but does it mean if i replace the console.log(i) which is returning the undefined, with a function returning the value of i . the code will work as expected ?
a-d
The first parameter of setTimeout must be a function or a string that will be evaled
Constantine
what about setTimeout(test(i),1000); function test(i) { return i; }
a-d
setTimeout(test(i),1000); will not work setTimeout(test.bind(null, i),1000); will work because bind returns a function, but does not call itfor(var i=0;i<5;i++) //this code loop for five time before it break.
i.e, it will execute this function for five time setTimeout(console.log(i),1000);.
setTimeout() is a javascript function for delay and it going to delayed for 1s and it going to carry out action for console.log(i);.
1 Comment
Constantine
i guess then it should make a delay 1000ms between logging the values
console.log()tosetTimeoutinstead of passing a function.