I've been trying to run the following code but only the second function runs and the first one does not execute. Can anyone please let me know what's wrong.
function first() {
setTimeout(function(){
console.log(1);
}, 500 );
};
function second(first) {
console.log(2);
};
second();What I'm expecting is that the program first displays 1 after 500ms and then 2.
What you expect :
function first() {
setTimeout(function(){
console.log(1);
second();
}, 500 );
};
function second() {
console.log(2);
};
first();
Your 'first' parameter in the second function does nothing. Instead you can do this too :
function first(callback) {
setTimeout(function(){
console.log(1);
callback();
}, 500 );
};
function second() {
console.log(2);
};
first(second);
I think this is the effect you're trying to achieve. first accepts a callback function as a parameter. It then needs to be called somewhere inside of the first function.
function first(callback) {
setTimeout(function() {
console.log(1);
callback(); // Calling the passed function
}, 500);
};
function second() {
console.log(2);
};
first(second); // Passing the 'second' function as a callback
console.log(1) before setTimeoutyou just called second. you most call first for callback funtion for you expectation your code shoould be like this
function second() {
setTimeout(function(){
console.log(2);
}, 500 );
};
function first() {
console.log(1);
second();
};
first();
