可选绑定试试？并作为？仍然会生成一个可选类型

问题描述

我有执行 NSFetchRequest 的代码，并将其结果转换为我的自定义数据模型类型的数组。提取可能会抛出，但我不想关心错误所以我使用尝试？，我还使用作为？在铸造中。在Swift 2中，这曾经很好，但是Swift 3产生了一个双重选项：

I have code for executing an NSFetchRequest and casting its result to an array of my custom data model type. Fetching may throw but I don't want to care about the error so I use try?, and I also use as? in casting. In Swift 2, this used to be just fine, but Swift 3 produces a double optional:

var expenses: [Expense]? {
    let request = NSFetchRequest<NSFetchRequestResult>(entityName: Expense.entityName)
    request.predicate = NSPredicate(format: "dateSpent >= %@ AND dateSpent <= %@", [self.startDate, self.endDate])

    // Returns [Expense]? because right side is [Expense]??
    if let expenses = try? App.mainQueueContext.fetch(request) as? [Expense],
        expenses?.isEmpty == false {
        return expenses
    }
    return nil
}

如果允许，我如何在中重新定义可选绑定的右侧，以便其类型简单是一个数组 [费用] ？我认为在下面的布尔条件（曾经是 where 子句）中看起来很荒谬，数组仍然是可选的。

How can I rephrase the right side of my optional binding in if let so that its type will simply be an array [Expense]? I think it looks absurd that in the following boolean condition (which used to be a where clause), the array is still optional.

推荐答案

你必须在括号内包装 try？，如下所示：

You must wrap your try? call within parenthesis like this :

if let expenses = (try? App.mainQueueContext.fetch(request)) as? [Expense]

那是因为为？有优先级高于尝试？（可能是因为 try？可以应用于整个表达式。）

That's because as? has a higher precedence than try? (probably because try? can be applied to the whole expression).

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