Question

我正在研究一个学校项目，一切正常，但在编译时，我得到警告＆＃34;指针和整数之间的比较＆＃34;。谁能解释为什么会这样？我在下面的所有四个共享内存初始化中收到相同的警告。我尝试了不同的方法来进行比较。据我所知，shmget和shmat应该在错误时返回-1。即使警告出现，代码也能很好地运行。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：提出建议编辑。标志比较不再抛出错误（枚举状态*）。现在我得到警告：不同指针类型的比较缺少其他三个（clockVar，shmMsg和pidList）的强制转换。我还在顶部添加了我的声明。我相信我只是搞砸了一些语法。谢谢！

Answer 1

shmat()会返回指针，您将其与整数进行比较。请参阅shmat() manpage：

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

更改您的代码（假设您的问题中未显示的变量声明具有正确的类型，在本例中为void *shmMsg;）：

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

你应该没事。

共享内存 - 警告：指针和整数之间的比较

1 个答案: