當我試圖找出我的應用程序是否在後臺時，我得到這兩個警告

Question

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

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

我得到這兩個警告。

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')) 

我不明白是什麼問題。我想知道，如果我的應用程序在後臺/非活動或前景

Answer 1

[application applicationState]返回一個值，而不是一個對象（或指向任何東西）。

嘗試：

UIApplicationState state = [application applicationState]; 

Answer 2

UIApplicationState是一個typedef定義枚舉，因此你不需要*。

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

您可以通過以下步驟修復您的代碼：

UIApplicationState state = [application applicationState]; 

Answer 3

UIApplicationState是一種原始數據類型是INT的typedef爲32位和長64位。 UIApplicationState使用NSInteger的數據類型的枚舉，並聲明它無需使用指針*在你的發言。