分配 struct file_operations 字段时从不兼容的指针类型初始化

Question

我正在编写一个 Linux 设备驱动器，我很困惑为什么会收到此警告。

error: initialization from incompatible pointer type [-Werror=incompatible-pointer-types]
  .write = file_write,

这是我创建的代码，当我使用 make 命令时出现上述错误。我尝试将 ssize_t 更改为 int 但我仍然遇到相同的错误。

#include <linux/module.h>
#include <linux/string.h>
#include <linux/fs.h>
#include <asm/uaccess.h>
#include <linux/random.h>

MODULE_LICENSE("NIKS");


static char msg[100]={0};
static short readPos =0;

static int file_open(struct inode * , struct file *);
static int file_release(struct inode * , struct file *);
static ssize_t file_read(struct file *, char *, size_t,loff_t * );
static ssize_t file_write(struct file *, char *, size_t,loff_t * );

static struct file_operations fo = 
{
    .read = file_read,
    .open = file_open,
    .write = file_write,
    .release = file_release,
};

int init_module(void)
{
    int t = register_chrdev(150,"encdev",&fo);
    if(t<0)
    {
        printk("error");
    }
    else
    {
        printk("success");
    }
    return t;
}

void cleanup_module(void)
{
    unregister_chrdev(150,"encdev");
}

static int file_open(struct inode *in , struct file *fil)
{
    return 0;
}

static ssize_t file_read(struct file *fil, char *buf, size_t len,loff_t *off )
{
    short count=0;
    printk("here %d",msg[0]!=0);
    while(len && (msg[readPos]!=0))
    {
        printk("read");
        put_user(msg[readPos],buf++);
        count++;
        len--;
        readPos++;
    }
    return count;
}

static ssize_t file_write(struct file *fil, char *buf, size_t len,loff_t *off )
{
    short ind = 0;
    short count =0;
    memset(msg,0,100);
    readPos = 0;
    int i =0;
    char bytes[16];
    char rand;
    while(i<16)
    {
        get_random_bytes(&rand, sizeof(rand));
        bytes[i]=rand;
        ++count;
        i++;
    }
    int _len = len+16;
    while(_len>0 || count%16){
        if(ind<16) msg[ind] = bytes[ind];
        else{
            msg[ind] = (_len>0?buf[ind-16]:'@')^msg[ind-16];
        }
        ++ind;
        --_len;
        ++count;
    }
    return count;

}

static int file_release(struct inode *in , struct file *fil)
{
    printk("done");
    return 0;
}

我是 C 语言的新手，我只是不明白为什么会收到此警告。任何人都可以解释为什么我收到此警告以及如何处理吗？

Answer 1

struct file_operations 的 read 和 write 函数指针的正确类型是：

ssize_t (*read) (struct file *, char __user *, size_t, loff_t *);
ssize_t (*write) (struct file *, const char __user *, size_t, loff_t *);

因此，在您的模块中，适当的签名是：

static ssize_t file_read(struct file *, char __user *, size_t, loff_t *);
static ssize_t file_write(struct file *, const char __user *, size_t, loff_t *);

// ...

static ssize_t file_read(struct file *fil, char __user *buf, size_t len,loff_t *off)
{

// ...

static ssize_t file_write(struct file *fil, const char __user *buf, size_t len, loff_t *off)
{

缺少的重要部分是 const 限定符，但同样重要的是要记住，从用户空间传递的任何指针都应该用 __user 注释标记（即使编译器不会如果您忘记了它，请警告您，它仅用作 Sparse).

的静态检查

你在其他语句之后还有几个声明，这在C90中是无效的。编译器会就这些警告你，按照警告将它们移到函数体的顶部。

分配 struct file_operations 字段时从不兼容的指针类型初始化

Initialization from incompatible pointer type when assigning struct file_operations fields

c

linux

kernel

kernel-module

linux-kernel