Question

有没有办法通过引用传递块？下面的代码打印出来＆＃34; Block是nil＆＃34;

- (BOOL)application:(UIApplication *)application didFinishLaunchingWithOptions:(NSDictionary *)launchOptions {

    void (^block)(void) = nil;
    [self assignBlock:block];

    if ( block ) {block();}
    else{NSLog(@"Block is nil");}

    return YES;
}

- (void)assignBlock:(void (^)(void))blockToAssign
{
    blockToAssign = ^(void){
        NSLog(@"Block assigned");
    };
}

Answer 1

是的，你可以这样做。声明一个类型：

typedef void (^MyBlock)();

重新定义接收指针的函数assignBlock：

- (void)assignBlock:(MyBlock *)blockToAssign
{
    // assign the object pointing, not the pointer
    *blockToAssign = ^(void){
        NSLog(@"Block assigned");
    };
}

现在使用类型

初始化块

MyBlock block = nil;

当你调用函数时，传递块的地址：

[self assignBlock:&block];

if ( block ) {block();}
else{NSLog(@"Block is nil");}

通过引用传递块

1 个答案: