You can see from below the body is executed immediately just by putting synchronous code in the body rather than asynchronous:
function doSomethingAsynchronous() {
return new Promise((resolve) => {
console.log("a");
resolve("promise result");
});
}
doSomethingAsynchronous();
console.log("b");
The result shows the promise body is executed immediately (before 'b' is printed).
The result of the Promise is retained, to be released to a 'then' call for example:
function doSomethingAsynchronous() {
return new Promise((resolve) => {
console.log("a");
resolve("promise result");
});
}
doSomethingAsynchronous().then(function(pr) {
console.log("c:" + pr);
});
console.log("b");
Result:
a
b
c:promise result
Same deal with asynchronous code in the body except the indeterminate delay before the promise is fulfilled and 'then' can be called (point c). So a and b would be printed as soon as doSomethingAsynchronous() returns but c appears only when the promise is fulfilled ('resolve' is called).
What looks odd on the surface once the call to then is added, is that b is printed before c even when everything is synchronous.
Surely a would print, then c and finally b?
The reason why a, b and c are printed in that order is because no matter whether code in the body is async or sync, the then method is always called asynchronously by the Promise.
In my mind, I imagine the then method being invoked by something like setTimeout(()=>{then(pr)},0) in the Promise once resolve is called. I.e. the current execution path must complete before the function passed to then will be executed.
Not obvious from the Promise specification why it does this?
My guess is it ensures consistent behavior regarding when then is called (always after current execution thread finishes) which is presumably to allow multiple Promises to be stacked/chained together before kicking off all the then calls in succession.