iphone - 我应该释放 NSURL 吗？

Question

我有以下代码：

    + (BOOL)updateStatus:(NSString *)status forUsername:(NSString *)username withPassword:(NSString *)password {

NSURL *loginURL = [NSURL URLWithString:@"XYZ"];
ASIFormDataRequest *loginRequest = [[ASIFormDataRequest requestWithURL:loginURL] retain]; 
[loginRequest setPostValue:@VALUE forKey:@"SOME_KEY"]; 
[loginRequest setPostValue:username forKey:@"username"]; 
[loginRequest setPostValue:password forKey:@"password"];  
[loginRequest startSynchronous]; 

int loginStatusCode = [loginRequest responseStatusCode];
[loginRequest release];

if (loginStatusCode == 200) { //if we were able to login
    NSURL *updateURL = [NSURL URLWithString:@"ABC"];
    ASIFormDataRequest *updateRequest = [[ASIFormDataRequest requestWithURL:updateURL] retain]; 
    [updateRequest setPostValue:@VALUE forKey:@"SOME_KEY"]; 
    [updateRequest setPostValue:VALUE   forKey:@"SOME_KEY"];
    [updateRequest setPostValue:@"VALUE" forKey:@"SOME_KEY"]; 
    [updateRequest startSynchronous]; 

    int statusCode = [updateRequest responseStatusCode];
    [updateRequest release];

    NSURL *logoutURL = [NSURL URLWithString:@"ABC"];
    ASIFormDataRequest *logoutRequest = [ASIFormDataRequest requestWithURL:logoutURL]; 
    [logoutRequest setPostValue:VALUE forKey:@"KEY"]; 
    [logoutRequest startSynchronous]; 

    [logoutRequest release];
    if (statusCode == 200) { /
        return YES;
    } else {
        return NO;
    }
} else {
    return NO; 
}
}

我在带有 Instruments 的模拟器上运行代码，它报告了以下泄漏： alt text http://img267.imageshack.us/img267/7651/instruments.png 指向以下行：

NSURL *loginURL = [NSURL URLWithString:@"XYZ"];

我很确定我不必发布 NSURL（由于约定我只需要发布使用 init 或 copy 创建的对象）。我错了吗？还是代码有其他问题？

Answer 1

据我所知，你是对的。您不需要释放 loginURL。

但是我注意到你在代码的其他地方做 overrelease logoutRequest。

// Note no retain of logoutRequest here as is done elsewhere
ASIFormDataRequest *logoutRequest = [ASIFormDataRequest requestWithURL:logoutURL];   
// ... 
[logoutRequest release];

Answer 2

为什么要发布 logoutRequest？你没有保留它。

我认为您对 NSURL 的使用没有任何问题，它们都是自动发布的。