为何向异步函数传递non-sendable类型未触发Sendable合规警告？

我实现了两个逻辑完全一致的类型：BankAccount（actor类型）和AlternativeBankAccount（普通类，未遵循Sendable协议）。但将AlternativeBankAccount实例传递给异步函数charge时，编译器并未触发任何警告。明明异步上下文从random方法切换到了charge函数，我原本以为这会强制要求charge的所有参数必须符合Sendable协议，希望有人能解释原因。

运行日志 #

1 - <_NSMainThread: 0x600001704040>{number = 1, name = main}

2 - <NSThread: 0x600001711100>{number = 7, name = (null)}

代码实现 #

actor BankAccount {
    enum BankError: Error {
        case insufficientFunds
    }

    var balance: Double

    init(initialDeposit: Double) {
        self.balance = initialDeposit
    }

    func withdraw(amount: Double) throws {
        guard balance >= amount else {
            throw BankError.insufficientFunds
        }
        balance -= amount
    }

    func deposit(amount: Double) {
        balance += amount
    }

}

class AlternativeBankAccount {
    enum BankError: Error {
        case insufficientFunds
    }

    var balance: Double

    init(initialDeposit: Double) {
        self.balance = initialDeposit
    }

    func withdraw(amount: Double) throws {
        guard balance >= amount else {
            throw BankError.insufficientFunds
        }
        balance -= amount
    }

    func deposit(amount: Double) {
        balance += amount
    }

}

struct Charger {
    func charge(amount: Double, from bankAccount: isolated BankAccount, to otherAccount: AlternativeBankAccount)
    async throws -> (Double, Double) {

        print("2 - \(Thread.current)")
        try bankAccount.withdraw(amount: amount)
        let newBalance = bankAccount.balance
        otherAccount.deposit(amount: amount)
        return (newBalance, otherAccount.balance)

    }
}

@MainActor
class ViewModel {

    let charger = Charger()

    func random(bankAccount: BankAccount) {

        let bankAccount2 = AlternativeBankAccount(initialDeposit: 200)

        Task {
            print("1 - \(Thread.current)")
            let aa = try? await charger.charge(amount: 100, from: bankAccount, to: bankAccount2)
            print("balance", aa?.0 ?? "", aa?.1 ?? "")
        }

    }
}

原因解析 #

1. 默认并发检查级别宽松：Swift默认的Strict Concurrency Checking设置为Minimal，这个级别下编译器不会严格检查非Sendable类型跨异步上下文/线程的传递。只有将该选项设置为Targeted或Complete时，才会触发相关警告。

2. 异步函数不强制参数为Sendable：异步函数本身不会自动要求所有参数都遵循Sendable协议，只有当参数需要在不同actor/任务之间传递（即跨线程访问）时，Sendable才是必须的——它用来保证类型的线程安全。你的代码中AlternativeBankAccount是引用类型，没有任何线程安全保障，跨线程传递会有数据竞争风险，但默认编译设置下编译器不会主动检测这个问题。

3. 如何触发警告：在Xcode的项目设置中，找到Build Settings -> Swift Compiler - Language -> Strict Concurrency Checking，将其改为Complete。此时编译器会明确提示AlternativeBankAccount未遵循Sendable协议，且无法安全跨线程传递。

内容的提问来源于stack exchange，提问作者Laura Corssac