“mmap 教程”不正确，还是 GCC 编译错误？

Question

这个 15 年前的 mmap tutorial 在 Google 搜索中排名靠前，但它实际上在我的 Linux 系统上运行不正确。

mmap_write.c:

#include <stdio.h>
#include <stdlib.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <unistd.h>
#include <fcntl.h>
#include <sys/mman.h>

#define FILEPATH "/tmp/mmapped.bin"
#define NUMINTS  (1000)
#define FILESIZE (NUMINTS * sizeof(int))

int main(int argc, char *argv[])
{
    int i;
    int fd;
    int result;
    int *map;  /* mmapped array of int's */

    /* Open a file for writing.
     *  - Creating the file if it doesn't exist.
     *  - Truncating it to 0 size if it already exists. (not really needed)
     *
     * Note: "O_WRONLY" mode is not sufficient when mmaping.
     */
    fd = open(FILEPATH, O_RDWR | O_CREAT | O_TRUNC, (mode_t)0600);
    if (fd == -1) {
    perror("Error opening file for writing");
    exit(EXIT_FAILURE);
    }

    /* Stretch the file size to the size of the (mmapped) array of ints
     */
    result = lseek(fd, FILESIZE-1, SEEK_SET);
    if (result == -1) {
    close(fd);
    perror("Error calling lseek() to 'stretch' the file");
    exit(EXIT_FAILURE);
    }
    
    /* Something needs to be written at the end of the file to
     * have the file actually have the new size.
     * Just writing an empty string at the current file position will do.
     *
     * Note:
     *  - The current position in the file is at the end of the stretched 
     *    file due to the call to lseek().
     *  - An empty string is actually a single '\0' character, so a zero-byte
     *    will be written at the last byte of the file.
     */
    result = write(fd, "", 1);
    if (result != 1) {
    close(fd);
    perror("Error writing last byte of the file");
    exit(EXIT_FAILURE);
    }

    /* Now the file is ready to be mmapped.
     */
    map = mmap(0, FILESIZE, PROT_READ | PROT_WRITE, MAP_SHARED, fd, 0);
    if (map == MAP_FAILED) {
    close(fd);
    perror("Error mmapping the file");
    exit(EXIT_FAILURE);
    }
    
    /* Now write int's to the file as if it were memory (an array of ints).
     */
    for (i = 1; i <=NUMINTS; ++i) {
    map[i] = 2 * i; 
    }

    /* Don't forget to free the mmapped memory
     */
    if (munmap(map, FILESIZE) == -1) {
    perror("Error un-mmapping the file");
    /* Decide here whether to close(fd) and exit() or not. Depends... */
    }

    /* Un-mmaping doesn't close the file, so we still need to do that.
     */
    close(fd);
    return 0;
}

mmap_read.c:

#include <stdio.h>
#include <stdlib.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <unistd.h>
#include <fcntl.h>
#include <sys/mman.h>

#define FILEPATH "/tmp/mmapped.bin"
#define NUMINTS  (1000)
#define FILESIZE (NUMINTS * sizeof(int))

int main(int argc, char *argv[])
{
    int i;
    int fd;
    int *map;  /* mmapped array of int's */

    fd = open(FILEPATH, O_RDONLY);
    if (fd == -1) {
    perror("Error opening file for reading");
    exit(EXIT_FAILURE);
    }

    map = mmap(0, FILESIZE, PROT_READ, MAP_SHARED, fd, 0);
    if (map == MAP_FAILED) {
    close(fd);
    perror("Error mmapping the file");
    exit(EXIT_FAILURE);
    }
    
    /* Read the file int-by-int from the mmap
     */
    for (i = 1; i <=NUMINTS; ++i) {
    printf("%d: %d\n", i, map[i]);
    }

    if (munmap(map, FILESIZE) == -1) {
    perror("Error un-mmapping the file");
    }
    close(fd);
    return 0;
}

如果文件不存在，则mmap_read的输出为

...
998: 1996
999: 1998
1000: 2000

但如果是的话，输出是

...
998: 1996
999: 1998
1000: 0

作者应该刷新write吗？还是 GCC 错误编译了代码？

---

编辑：我注意到文件的先前存在或不存在会有所不同，而不是编译标志。

Answer 1

您从第二个元素开始，并在地图结束后写入 2000。

for (i = 1; i <=NUMINTS; ++i) {
    map[i] = 2 * i; 
}

应该是

for (i = 0; i < NUMINTS; ++i) {
    map[i] = 2 * ( i + 1 ); 
}

Demo

这不是缓冲问题。 write 是系统调用，所以数据直接传递给操作系统。这并不意味着当write 返回时数据已写入磁盘，但它在操作系统手中，因此就操作系统功能而言，就好像它在磁盘上，包括它的内存映射功能。

Answer 2

在 C 中，索引从零开始。写和读索引 1000 你调用了未定义的行为

改写为：

   for (i = 1; i <=NUMINTS; ++i) {
       map[i - 1] = 2 * i; 
   }

并阅读：

    for (i = 1; i <=NUMINTS; ++i) {
    printf("%d: %d\n", i, map[i-1]);
    }