Objective-c中的Smalltalk块？

Question

Objective-C是否支持阻止"a la Smalltalk"？

在Smalltalk中,块类似于其他语言中的"闭包"或"lambda表达式"或"无名函数".

Answer 1

是的:http://developer.apple.com/library/mac/#documentation/Cocoa/Conceptual/Blocks/Articles/00_Introduction.html

开箱即用,它们仅在XCode 3.2或更高版本附带的Objective-C 2.0版本中得到支持.这意味着如果您想使用官方构建工具,您将需要Snow Leopard.这里描述了10.5的潜在解决方法:http://thirdcog.eu/pwcblocks/#leoiphone

Answer 2

是的,举个例子:

[[myString componentsSeparatedByString:@"\n"] enumerateObjectsWithOptions:NSEnumerationConcurrent usingBlock:^(id obj, NSUInteger idx, BOOL *stop) {
    NSString *line = (NSString *)obj;
    //do what your going to do with line...
}];