UNIX阅读行为 [英] UNIX read behavior

问题描述

我写了这个小code，以确定读的行为。

I wrote this small code to ascertain read behavior. 

#include <stdio.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>
#include <limits.h>
#include <errno.h>

int main ()
{
  ssize_t ret;
  int fd;
  char str[30] = {0};
  off_t lret

  fd = open("./sample", O_RDWR);
  printf("File descriptor = %d\n",fd);

  lret = lseek(fd,LONG_MAX,SEEK_SET);
  printf("%ld\n",lseek(fd, 0, SEEK_CUR));

  ret = read(fd, str, 20);
  if (ret == -1) {
     perror("read error");
  }
  else {
     printf("%ld\n",ret);
     printf("%s\n",str);
  }

  ret = write(fd, "bye", 3);
  if (ret == -1) {
     perror("write error");
  }
  else
     printf("%ld\n",ret);

  printf("%ld\n",lseek(fd, 0, SEEK_CUR));
  close (fd);

  return 0;
}

下面是输出：

$ cat sample
HELLO$ ./a.out
File descriptor = 3
4294967295
read error: Invalid argument
write error: Invalid argument
4294967295
$ ll sample
-rw-r--r--. 1 bruce stud 5 Jan 14 17:25 sample

但是，如果我更改lseek的语句

But if I change lseek statement to

ret = lseek(fd,5,SEEK_SET);

read返回0

read returns 0

$ ./a.out
File descriptor = 3
5
0

3
8
$ cat sample
HELLObye$ ll sample
-rw-r--r--. 1 bruce stud 8 Jan 14 17:26 sample

为什么读这样的表现？

Why does read behave like this?

推荐答案

请注意，通过 lseek的返回的值是一个 off_t ，而不是为size_t 。所不同的是off_t签署。当你把一个符号值，使之无符号，它看起来像一个大的正数。

Note that the value returned by lseek is a off_t, not a size_t. The difference is that off_t is signed. When you take a signed value and make it unsigned, it appears like a large positive number.

我预计LONG_MAX其实不是 4294967295 ，但无论是 2147483647 （2 ^ 31-1 ），或者一个更大的数。因此 4294967295 来自-1 [它是2 ^ 32-1，这确实是在32位的运算一样-1]。

I would expect that "LONG_MAX" is not actually 4294967295, but either 2147483647 (2^31-1) or a much larger number. So the 4294967295 comes from a -1 [it is 2^32-1, which is indeed the same as -1 in 32-bit math].

在换句话说，你都可以从 lseek的错误。

In other words, you are getting an error from lseek.

这篇关于UNIX阅读行为的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！