In the following function
function test1(n, delay) {
for (let i = 0; i < n; i++) {
setTimeout(() => {
console.log(i)
}, delay)
}
}
test1(3, 1000)
After 1 second, I immediately console.log 1, 2, and 3 simultaneously.
If I multiply delay by i,
function test1(n, delay) {
for (let i = 0; i < n; i++) {
setTimeout(() => {
console.log(i)
}, i * delay)
}
}
test1(3, 1000)
I console.log every i in my loop after 1 second. Why does this work?
Also, why does the code below
function test2(n, delay) {
let promise = new Promise(resolve => setTimeout(() => {
resolve()
}, delay))
for (let i = 0; i < n; i++) {
promise.then(console.log(i))
}
}
test2(3, 1000)
immediately console.log 1, 2, and 3 simultaneously instead of waiting every second to console.log the next i value? Is there a way to console.log the next i value after waiting every second without using async and await?
