You asked:
will first task always finish before second task starts?
No, it will not. Whenever you see await
, that is a “suspension point” at which Swift concurrency is free to switch to another task. In short, these can run concurrently. Let me illustrate that with Xcode Instruments:
import os.log
private let log = OSLog(subsystem: "Test", category: .pointsOfInterest)
class Foo {
func fetchImages() {
Task {
let fetch = await loadImages()
print("done loadImages")
}
Task {
let fetch1 = await loadImages1()
print("done loadImages1")
}
}
func loadImages() async {
// start “points of interest” interval
let id = OSSignpostID(log: log)
os_signpost(.begin, log: log, name: #function, signpostID: id, "start")
// perform 3-second asynchronous task
try? await Task.sleep(nanoseconds: 3 * NSEC_PER_SEC)
// end “points of interest” interval
os_signpost(.end, log: log, name: #function, signpostID: id, "end")
}
func loadImages1() async {
// start “points of interest” interval
let id = OSSignpostID(log: log)
os_signpost(.begin, log: log, name: #function, signpostID: id, "start")
// perform 1-second asynchronous task
try? await Task.sleep(nanoseconds: 1 * NSEC_PER_SEC)
// end “points of interest” interval
os_signpost(.end, log: log, name: #function, signpostID: id, "end")
}
}
Profiling (in Xcode, either press command-i or choose from the menu, “Product” » “Profile”) this with the “time profiler” in Instruments, you can see that these run concurrently:

The trick is to have the second task await
the first one. E.g.,
func fetchImages() {
let firstTask = Task {
let fetch = await loadImages()
print("done loadImages")
}
Task {
_ = await firstTask.result
let fetch1 = await loadImages1()
print("done loadImages1")
}
}
Or you can store the first task in some property:
var firstTask: Task<Void, Never>? // you may need to adjust the `Success` and `Failure` types to match your real example
func fetchImages() {
firstTask = Task {
let fetch = await loadImages()
print("done loadImages")
}
Task {
_ = await firstTask?.result
let fetch1 = await loadImages1()
print("done loadImages1")
}
}
When you do that, you can see the sequential execution:

FWIW, this concept, of using await
on the prior task was the motivating idea behind that other answer that you referenced. That is a more generalized rendition of the above. Hopefully this illustrates the mechanism outlined in that other answer.