当我试图查明我的应用程序是否在后台时，我收到了这两个警告

Question

UIApplicationState *state = [application applicationState];
if(state == UIApplicationStateActive)
{
    NSLog(@"Display UIAlert");
}

if((state == UIApplicationStateBackground)||(state == UIApplicationStateInactive))
{
   NSLog(@"App is in background");
}

我收到这两个警告。

Incompatible integer to pointer conversion initializing 'UIApplicationState *' (aka 'enum UIApplicationState *') with an expression of type 'UIApplicationState' (aka 'enum UIApplicationState')

Comparison between pointer and integer ('UIApplicationState *' (aka 'enum UIApplicationState *') and 'NSInteger' (aka 'long'))

我不明白这是什么问题。我想知道我的应用程序是在 background/inactive 还是前台

Answer 1

[application applicationState] returns 一个值，而不是一个对象（或指向任何东西的指针）。

尝试：

UIApplicationState state = [application applicationState];

Answer 2

UIApplicationState 是一个类型定义的枚举，因此您不需要 *.

typedef enum : NSInteger {
   UIApplicationStateActive,
   UIApplicationStateInactive,
   UIApplicationStateBackground 
} UIApplicationState;

您可以通过执行以下操作来修复您的代码：

UIApplicationState state = [application applicationState];

Answer 3

UIApplicationState 是一种原始数据类型，32 位类型定义为 int，64 位类型定义为 long。 UIApplicationState 使用 NSInteger 数据类型的枚举，声明它不需要在你的语句中使用指针*。

当我试图查明我的应用程序是否在后台时，我收到了这两个警告

I am getting these two warnings when trying to find out if my application is in the background

objective-c

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