RxJava的observeOn()和subscribeOn()調用顺序是否有影响

[ws_lm]

下面observeOn()在subscribeOn()之前是不会有输出的，但是顺序换一下有输出，这是什么原因？
[code=java][Subscriber<String> observer1=new Subscriber<String>() {
@Override
public void onCompleted() {
System.out.println("onComplete");
}

@Override
public void onError(Throwable e) {
System.out.println("error");
}

@Override
public void onNext(String s) {
System.out.println("onNext");
}
};


Observable.create(new Observable.OnSubscribe<String>() {
@Override
public void call(Subscriber<? super String> subscriber) {
System.out.println("call:" + Thread.currentThread().getName());
subscriber.onNext("");
subscriber.onCompleted();
}
}).observeOn(Schedulers.newThread()).subscribeOn(Schedulers.newThread()).subscribe(observer1);/code]

[afterward__]

observeOn对下有影响。subscribeOn对上有影响。并列写的话两个顺序没有关系吧