async function f() {
return 42;
}
f().then(console.log);
the output of the function is 42
I have not much experience, but without an await
I don’t think you actually have asynchronous code
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I am not sure, but i think if you use .then function, it will perform as a sync function.
check it by your self.(you don’t need to use await function, if you use then func.)
An async function returns a Promise. You are calling .then() on the Promise and passing console.log as the callback for the onFulfilled
handler.
async function f() {
return 42;
}
// console.log as callback
f().then(console.log);
// the above is the same as explicitly passing the value
f().then((value) => console.log(value));
// Or passing your own callback
function logValue(value) {
console.log(value);
}
f().then(logValue);
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