I know 2 will be console.logged first and then 1. Is there any way, maybe with callbacks, I can show 1 first and then 2?
function first(){
// Simulate a code delay
setTimeout( function(){
console.log(1);
}, 500 );
}
function second(){
console.log(2);
}
first();
second();
Put second()'s call in the setTimeout:
function first(){
// Simulate a code delay
setTimeout( function(){
console.log(1);
second();
}, 500 );
}
function second(){
console.log(2);
}
first();
You could also bypass first, second and setTimeout completely:
console.log(1);
console.log(2);
function second(){
console.log(2);
}
function first(cb){
// Simulate a code delay
return setTimeout( function(){
console.log(1);
cb();
}, 500 );
}
first(second);Solve it by promise a clan and elegant way:
function second() {
console.log("test2", 2);
}
function first() {
var promise = new Promise(function (resolve, reject) {
setTimeout(function () {
console.log("test1", 1);
resolve();
}, 500);
});
return promise;
}
first().then(function () {
second();
});don't use timeout call in first function. or you can use promises
for better understanding of event loop callback execution order read this article
link. this article explains execution order , which get priority among promise , timeout etc
Hope it will help
second()in timeoutfirstorsecond; just doconsole.log(1); console.log(2);.firstfunction