Question

if(!setFlag)
{
    Intent myIntent = new Intent(FirstActivity.this, LogoutService.class);
    pendingIntent = PendingIntent.getService(FirstActivity.this, 0, myIntent, 0);
    AlarmManager alarmManager = (AlarmManager)getSystemService(ALARM_SERVICE);
    Calendar calendar = Calendar.getInstance();
    calendar.setTimeInMillis(System.currentTimeMillis());
    calendar.add(Calendar.MINUTE, 10); //Minutes
    alarmManager.set(AlarmManager.RTC_WAKEUP, calendar.getTimeInMillis(), pendingIntent);
}

我是否理解正确，只有a和b之间的区别是别名的创建＆＃34; v＆＃34;为＆＃34; vec＆＃34;在b情况下还是有更多的东西？

Answer 1

在这种情况下，没有真正的区别。但是，如果您按值捕获，则会有所不同：

const std::vector<int> vec; // note const

auto foo = [vec]() mutable {
   // can't change vec here since it is captured with cv-qualifiers
};

auto bar = [v = vec]() mutable {
    // can change v here since it is captured by auto deduction rules
    // (cv-qualifiers dropped)
};

Answer 2

基本上没有。它只是允许您为捕获的值提供自定义名称。

但是请注意，当您想要将对象实际移动到闭包中时，需要使用此语法（称为init捕获）：

std::vector<int> vector;
auto lambda = [vec = std::move(vector)]() { /* the lambda owns the vector now */ });

在[＆amp; capture]和[＆amp; local = capture]在lambdas中捕获之间有什么区别？

2 个答案: