如果代码跳过宏任务，回调 function 的目的是什么？

Question

Recently, I've tried to understand callback functions in JavaScript, however, this concept is still far away from my understanding. I have code like this:

function exampleFunc(callback) {
    console.log("Starting...");
    setTimeout(() => {
        console.log("Logged after 3 secs.");
    }, 3000);
    callback();
}

function displayMessage() {
    console.log("Log it!");
}
    
exampleFunc(() =>
{
    console.log("Further code.");
});

displayMessage();

I've expected that after calling exampleFunc() , program will wait 3 seconds, and then call the callback and rest of code. But instead, the sequence of code is:

Starting...
Further code.
Log it!
Logged after 3 secs.

Why does it happen? I've expected that program will output "Starting" , then wait and log "Logged after 3 secs." , then go to callback and output "Further code" , and in the end, "Log it!" .

Answer 1

Calling a setTimeout does not cause further execution of the code in that function to delay. setTimeout schedules a timeout, then continues executing the rest of the code in the function immediately .

You need to call callback inside the setTimeout callback, so that it runs only after the 3 seconds are up.

You also need to put the call of displayMessage inside the callback passed to exampleFunc .

 function exampleFunc(callback) { console.log("Starting..."); setTimeout(() => { console.log("Logged after 3 secs."); callback(); }, 3000); } function displayMessage() { console.log("Log it;"). } exampleFunc(() => { console.log("Further code;"); displayMessage(); });