我想知道allOf方法是否CompletableFuture進(jìn)行輪詢或進(jìn)入等待狀態(tài)，直到所有CompletableFutures傳遞給該方法的方法都完成執(zhí)行。allOf我查看了方法的代碼IntelliJ，它正在進(jìn)行某種二進(jìn)制搜索。請幫助我找出方法allOf的CompletableFuture實(shí)際作用。public static CompletableFuture<Void> allOf(CompletableFuture<?>... cfs) {    return andTree(cfs, 0, cfs.length - 1);}/** Recursively constructs a tree of completions. */static CompletableFuture<Void> andTree(CompletableFuture<?>[] cfs, int lo, int hi) {    CompletableFuture<Void> d = new CompletableFuture<Void>();    if (lo > hi) // empty        d.result = NIL;    else {        CompletableFuture<?> a, b;        int mid = (lo + hi) >>> 1;        if ((a = (lo == mid ? cfs[lo] :                  andTree(cfs, lo, mid))) == null ||            (b = (lo == hi ? a : (hi == mid+1) ? cfs[hi] :                  andTree(cfs, mid+1, hi)))  == null)            throw new NullPointerException();        if (!d.biRelay(a, b)) {            BiRelay<?,?> c = new BiRelay<>(d, a, b);            a.bipush(b, c);            c.tryFire(SYNC);        }    }    return d;}/** Pushes completion to this and b unless both done. */final void bipush(CompletableFuture<?> b, BiCompletion<?,?,?> c) {    if (c != null) {        Object r;        while ((r = result) == null && !tryPushStack(c))            lazySetNext(c, null); // clear on failure        if (b != null && b != this && b.result == null) {            Completion q = (r != null) ? c : new CoCompletion(c);            while (b.result == null && !b.tryPushStack(q))                lazySetNext(q, null); // clear on failure        }    }}final CompletableFuture<V> tryFire(int mode) {    CompletableFuture<V> d;    CompletableFuture<T> a;    CompletableFuture<U> b;    if ((d = dep) == null ||        !d.orApply(a = src, b = snd, fn, mode > 0 ? null : this))        return null;    dep = null; src = null; snd = null; fn = null;    return d.postFire(a, b, mode);}