NSBlockOperation和块中的对象

Question

下面是简单的代码：

// let's assume that I have to allocate this variable with alloc/init
NSString *someString = [[NSString alloc] initWithFormat:"%@", @"someString"];

NSBlockOperation *op = [NSBlockOperation blockOperationWithBlock:^{
  [someClass someFunction: someString];
}];

[queue addOperation:op]

[someString release];

在someString发布后运行released操作时，此代码将崩溃。防止这种情况的最佳做法是什么？

谢谢。

编辑:弧不是一个选择，因为这不是我的决定。在MRC有办法绕过这件事吗？

EDIT2:那么下面的代码呢？能行吗？

// let's assume that I have to allocate this variable with alloc/init
NSString *someString = [[NSString alloc] initWithFormat:"%@", @"someString"];

[someString retain]
NSBlockOperation *op = [NSBlockOperation blockOperationWithBlock:^{
  [someClass someFunction: someString];
  [someString release]
}];

[queue addOperation:op]

[someString release];

Answer 1

您确实应该使用自动引用计数，并将代码简化为

// let's say the variable is allocated with alloc/init
NSString *someString = @"someString";

NSBlockOperation *op = [NSBlockOperation blockOperationWithBlock:^{
  [someClass someFunction: someString];
}];

[queue addOperation:op]

如果您真的必须使用手动参考计数，您可以这样做：

// let's assume that I have to allocate this variable with alloc/init
NSString *someString = [[NSString alloc] initWithFormat:"%@", @"someString"];

NSBlockOperation *op = [NSBlockOperation blockOperationWithBlock:^{
  [someClass someFunction: someString];
  [someString release]
}];

[queue addOperation:op]

我知道这只是个示例代码，但如果不是，你也可以这么做.;)

NSBlockOperation *op = [NSBlockOperation blockOperationWithBlock:^{
  [someClass someFunction:@"someString"];
}];

[queue addOperation:op]

Answer 2

// let's say the variable is allocated with alloc/init
NSString *someString = [[[NSString alloc] initWithFormat:"%@", @"someString"] autorelease];

NSBlockOperation *op = [NSBlockOperation blockOperationWithBlock:^{
  [someClass someFunction: someString];
}];

[queue addOperation:op];

有成千上万的应用程序需要维护和不使用ARC。将字符串包装在自动发行版中。我认为这应该是可行的，但我没有测试。