Question

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

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

我收到了这两个警告。

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'))

我不明白这个问题是什么。我想知道我的应用程序是在后台/不活动还是前台

Answer 1

UIApplicationState是一个typedef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代码：

UIApplicationState state = [application applicationState];

Answer 2

[application applicationState]返回一个值，而不是一个对象（或指向任何东西的指针）。

尝试：

UIApplicationState state = [application applicationState];

Answer 3

UIApplicationState是一种原始数据类型，对于32位是int的typedef而对于64位是long。 UIApplicationState使用NSInteger数据类型的枚举，并声明它不需要在语句中使用指针*。

当我试图找出我的应用程序是否在后台时，我收到了这两个警告

3 个答案: