如何从文件描述符获得内核空间的绝对路径

Question

我想挂钩unlinkat.my挂钩函数。

但我只获取文件名而不是绝对路径.so我想绝对路径来比较string.when我尝试rm -r比我只获得文件名如果我获得绝对路径然后它work.so请告诉我如何我得到绝对路径。
我的代码是

long mw_sys_unlink(int dfd, const char *filename ,int flag) 
{ 
     long ret; 
     if(strstr(filename,"/tmp/a/")) 
     { 
       printk(KERN_INFO "file %s has not been deleted by kernel module\n", filename); 
       return -1; 
     } 
     else 
     { 
       ret = orig_sys_unlink(dfd ,filename,flag); 
       printk(KERN_INFO "file %s has been deleted", filename); 
       return ret; 
     } 
} 

Answer 1

尝试以下操作：

char *tmp = (char*)__get_free_page(GFP_TEMPORARY); 

    file *file = fget(dfd); 
    if (!file) { 
     goto out 
    } 

    char *path = d_path(&file->f_path, tmp, PAGE_SIZE); 
    if (IS_ERR(path)) { 
     printk("error: %d\n", (int)path); 
     goto out; 
    } 

    printk("path: %s\n", path); 
out: 
    free_page((unsigned long)tmp);