无法修改函数体内的函数参数

Question

我在swift项目中有一个方法定义:

class func fireGetRequest(urlString: String!, username: String?, password: String?, completionBlock:(NSDictionary)->Void) {
    //check if user passed nil userName
    if username == nil || password == nil {
        // retrieve userName, password from keychain
        // here we have OR check since we have single value from pair it is of no use and can be considered as corrupted data
        // so it is better to retrieve stable data from keychain for further processing
        let (dataDict, error) = Locksmith.loadDataForUserAccount(kKeychainUserAccountName)

        // no error found :)
        // use data retrieved
        if error == nil {
            username = dataDict[kKeychainUserNameKey]
            password = dataDict[kKeychainUserPwdKey]
        }
    }

    // do something with username, password
}

我在这里做以下事情:

1. 检查用户名,密码是否为零
2. 如果其中任何一个为零,则尝试从字典中设置相应的值

问题是 - 我无法解决以下错误:

objective-c中的类似方法完美地起作用:

+ (void)fireGetRequestWithUrlString:(NSString *)urlString userName:(NSString *)username userPwd:(NSString *)password{
    NSDictionary *dataDict = @{kKeychainUserNameKey: @"Some user name", kKeychainUserPwdKey: @"Some password"};

    if (!username || !password) {
        username = dataDict[kKeychainUserNameKey];
        password = dataDict[kKeychainUserPwdKey];
    }

    // do something with username, password
}

有任何想法吗？

更新1:

正如答案所示,我将函数参数 - 用户名和密码声明为var,但后来我开始收到这些编译错误:

为了解决这个问题,我确实输入了类型,但又出现了另

关于力量向下转发,低于错误:

仍然无能为力:(

终于解决了:)

if let dataDictionary = dataDict {
                    username = dataDictionary[kKeychainUserNameKey] as! String?
                    password = dataDictionary[kKeychainUserPwdKey] as! String?
                }

Answer 1

默认情况下,函数参数是常量.

明确定义可变参数为 var

斯威夫特2

class func fireGetRequest(urlString: String!, var username: String?, var password: String?, completionBlock:(NSDictionary)->Void) 

斯威夫特3

class func fireGetRequest(urlString: String!, username: String?, password: String?, completionBlock:(NSDictionary)->Void){
   var username = username
   var password = password
   //Other code goes here
}