共享内存 - 警告：指针和整数之间的比较

Question

我正在学校项目上工作，一切正常，但在编译时，我得到了“指针和整数之间的比较”警告。谁能解释为什么会发生这种情况？我在下面的所有四个共享内存初始化中都得到相同的警告。我尝试了不同的方法进行比较。据我所知，shmget和shmat在错误时应该返回-1。即使警告出现了，代码也能很好地工作。我做了更改以提示错误。这些警告在shmat线上提示。下面列出的最小代码。

//-------------------------------------------------- 
//Variables 
//-------------------------------------------------- 

//Shared memory variables 
int (*clockVar)[2]; 
int *turn; 
int (*shmMsg)[2]; 
enum state {idle, want_in, in_cs, done, dne} *flag; 
long *pidList; 

//Shared memory keys 
key_t clockKey; 
key_t turnKey; 
key_t msgKey; 
key_t flagKey; 
key_t pidKey; 

//Shared memory IDs 
int clockID = 0; 
int turnID = 0; 
int msgID = 0; 
int flagID = 0; 
int pidID = 0; 
//-------------------------------------------------- 
//Key Initialization 
//-------------------------------------------------- 

clockKey = ftok("ftok_clock", 13); 
if (clockKey == -1){ 
    perror("Clock: Failed to load ftok file"); 
    return 1; 
} 

msgKey = ftok("ftok_msg", 17); 
if (msgKey == -1){ 
    perror("Message: Failed to load ftok file"); 
    return 1; 
} 

flagKey = ftok("ftok_flag", 15); 
if (flagKey == -1){ 
    perror("Flag: Failed to load ftok file"); 
    return 1; 
} 

pidKey = ftok("ftok_pids", 17); 
if (pidKey == -1){ 
    perror("PID: Failed to load ftok file"); 
    return 1; 
} 

//-------------------------------------------------- 
//Shared Memory Initialization 
//-------------------------------------------------- 

//Initializing shared memory for clock counter 
clockID = shmget(clockKey, sizeof(int[2][1]), IPC_CREAT | 0666); 
if (clockID == -1){ 
    perror("Clock: Failed to designate shared memory"); 
    return 1; 
} 

clockVar = shmat(clockID, 0, 0); 
if (clockVar == (int*)-1){ 
    perror("Clock: Failed to attach shared memory"); 
    return 1; 
} 

//Initializing shared memory for the Message array 
msgID = shmget(msgKey, sizeof(int[2][1]), IPC_CREAT | 0666); 
if (msgID == -1){ 
    perror("Message: Failed to designate shared memory"); 
    return 1; 
} 

shmMsg = shmat(msgID, NULL, 0); 
if (shmMsg == (int*)-1){ 
    perror("Message: Failed to attach shared memory"); 
    return 1; 
} 

//Initializing shared memory for flag counter 
flagID = shmget(flagKey, sizeof(enum state[maxProc]), IPC_CREAT | 0666); 
if (flagID == -1){ 
    perror("Flag: Failed to designate shared memory"); 
    return 1; 
} 

flag = shmat(flagID, NULL, 0); 
if (flag == (enum state*)-1){ 
    perror("Flag: Failed to attach shared memory"); 
    return 1; 
} 

//Initializing shared memory for the process count array 
pidID = shmget(pidKey, sizeof(int), IPC_CREAT | 0666); 
if (pidID == -1){ 
    perror("PID: Failed to designate shared memory"); 
    return 1; 
} 

pidList = shmat(pidID, NULL, 0); 
if (pidList == (int*)-1){ 
    return 1; 
} 

编辑：作出建议编辑。标志比较不会再抛出错误（枚举状态*）。现在我得到警告：不同指针类型的比较缺乏对其他三个（clockVar，shmMsg和pidList）的强制转换。我还将我的声明添加到顶部。我相信我只是搞乱了一些语法。谢谢！

Answer 1

shmat()返回一个指针，并且您将它与一个整数进行比较。请参阅shmat() manpage：

成功时shmat()返回附加共享内存段的地址;在错误(void *) -1返回

更改您这样的代码（假设你的变量声明，你不要在你的问题显示有正确的类型，在这种情况下，void *shmMsg;）：

// [...] 
shmMsg = shmat(msgID, NULL, 0); 
if (shmMsg == (void*) -1){ 
    //[...] 

和你应该没事。