Question

我正在尝试了解NSOperationQueue，并尝试创建最简单的示例。我有以下内容：

NSOperationQueue *myOQ=[[NSOperationQueue alloc] init];

[myOQ addOperationWithBlock:^(void){
  NSLog(@"here is something for jt 2");
}];
[myOQ addOperationWithBlock:^(void){
  NSLog(@"oh is this going to work 2");
}];

但是想这样做：

void  (^jt)() = ^void(){
  NSLog(@"here is something for jt");
};

void (^cl)() = ^void(){
  NSLog(@"oh is this going to work");
};

NSOperationQueue *myOQ=[[NSOperationQueue alloc] init];

[myOQ addOperation:jt];
[myOQ addOperation:cl];

后一种形式可能吗？我可以将块转换为NSOperation吗？

事先提前

Answer 1

你可以：

NSBlockOperation *jtOperation = [NSBlockOperation blockOperationWithBlock:^{
    NSLog(@"here is something for jt");
}];

NSBlockOperation *clOperation = [NSBlockOperation blockOperationWithBlock:^{
    NSLog(@"oh is this going to work");
}];

[myOQ addOperation:jtOperation];
[myOQ addOperation:clOperation];

话虽如此，我通常会addOperationWithBlock，除非我出于其他原因需要NSOperation对象指针（例如，建立操作之间的依赖关系等）。

Answer 2

您也可以

[operationQueue addOperationWithBlock:^{
    // Stuff
})];

Answer 3

<强>夫特

let networkingOperation = NSBlockOperation(block: {
                             // Your code here
                          })

有没有办法像这样向NSOperationQueue添加块

3 个答案: