使用宏定义替代控制器跳转

1、导航push
宏定义

#define NAVPUSH(_VIEWCONTROLLER_) ({\ _VIEWCONTROLLER_ *(^block)(void) = ^{_VIEWCONTROLLER_ *controller = [[_VIEWCONTROLLER_ alloc]init];\ [self.navigationController pushViewController:controller animated:true];\ return controller;};\ LeJSZDViewController*controller = block();\ (controller);\ })\         

使用方法触发push效果 方便调用一句话实现push 也可以穿参

NAVPUSH(LeJSZDViewController).name = @"极速指导";   //传参数
NAVPUSH(LeJSZDViewController)//也可以不穿参数
LeJSZDViewController *vc = NAVPUSH(LeJSZDViewController)；//也可以获取对象

以前push是这样的

LeJSZDViewController *controller = [[LeJSZDViewController alloc]init];
 controller.name = @"极速指导";
 [self.navigationController pushViewController:controller animated:true];

使用宏之后是这样的

NAVPUSH(LeJSZDViewController);    

如果要传参数(只有一个参数)

NAVPUSH(LeJSZDViewController).name = @"极速指导"; 

有多个参数

LeJSZDViewController * controller = NAVPUSH(LeJSZDViewController);     
controller.name = @"极速指导";
controller.title = @"极速指导";

2.STORYBOARD便捷引用

#define NAVPUSHSTORYBOARD(_STORYBOARDVIEWCONTROLLER_,_NAME_,_INDENTIFER_) ({\
_STORYBOARDVIEWCONTROLLER_ *(^block)(void) = ^{_STORYBOARDVIEWCONTROLLER_ *controller = [[UIStoryboard storyboardWithName:_NAME_ bundle:nil] instantiateViewControllerWithIdentifier:_INDENTIFER_];\
[self.navigationController pushViewController:controller animated:true];\
return controller;};\
_STORYBOARDVIEWCONTROLLER_*controller = block();\
(controller);\
})\

3.模态

#define PRENT(_VIEWCONTROLLER_)({\
    _VIEWCONTROLLER_ *(^block)(void) = ^{_VIEWCONTROLLER_ *controller = [[_VIEWCONTROLLER_ alloc]init];\
        [self presentViewController:controller animated:true completion:nil];\
        return controller;};\
    _VIEWCONTROLLER_*controller = block();\
    (controller);\
})\

2和3 使用方法和1 一样 个人觉得比较简单，分享！！！