Question 213488 in Javascript Question posted in
A very good W3school tutorial can be found here.

Javascript – Why does the second "then" logs "OK" instead of "undefined"?

var p = new Promise(function(resolve, reject) {
    setTimeout(function() {
        resolve("OK");
    }, 2000);
});

p.then().then(function(data) {
    console.log(data);
});

I don’t understand what happens after the first "then" is called. When the first "then" is called it returns undefined, which in turn resolves Promise created for this "then" with a value "undefined". Yet, for some reason, when the second "then" is called it is passed "OK" as "data" and subsequently logs it. Where did the "data" come from ?

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2

Answers


  1. 0

    This "forwarding" of the value is intended behaviour. Mozilla contributors write:

    Parameters

    onFulfilled Optional

    A function to asynchronously execute when this promise becomes fulfilled.
    […]

    If it is not a function, it is internally replaced with an identity function ((x) => x) which simply passes the fulfillment value forward.

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  2. 0 
    var p = new Promise(function(resolve, reject) {
    setTimeout(function() {
        resolve("OK");
    }, 2000);
});

p.then(console.log); // prints ok
p.then().then().then().then(console.log); // prints ok
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