关于 Linux 直接 I/O 的问题

在看《 The Linux Programming Interface 》（ Linux/Unix 系统编程手册）第 13 章，第 6 小节，其中的示例代码:

/* direct_read.c Demonstrate the use of O_DIRECT to perform I/O bypassing the buffer cache ("direct I/O"). Usage: direct_read file length [offset [alignment]] This program is Linux-specific. */ #define _GNU_SOURCE /* Obtain O_DIRECT definition from <fcntl.h> */ #include <fcntl.h> #include <malloc.h> #include "tlpi_hdr.h" int main(int argc, char *argv[]) { int fd; ssize_t numRead; size_t length, alignment; off_t offset; char *buf; if (argc < 3 || strcmp(argv[1], "--help") == 0) usageErr("%s file length [offset [alignment]]\n", argv[0]); length = getLong(argv[2], GN_ANY_BASE, "length"); offset = (argc > 3) ? getLong(argv[3], GN_ANY_BASE, "offset") : 0; alignment = (argc > 4) ? getLong(argv[4], GN_ANY_BASE, "alignment") : 4096; fd = open(argv[1], O_RDONLY | O_DIRECT); if (fd == -1) errExit("open"); /* memalign() allocates a block of memory aligned on an address that is a multiple of its first argument. By specifying this argument as 2 * 'alignment' and then adding 'alignment' to the returned pointer, we ensure that 'buf' is aligned on a non-power-of-two multiple of 'alignment'. We do this to ensure that if, for example, we ask for a 256-byte aligned buffer, we don't accidentally get a buffer that is also aligned on a 512-byte boundary. */ buf = memalign(alignment * 2, length + alignment); if (buf == NULL) errExit("memalign"); buf += alignment; if (lseek(fd, offset, SEEK_SET) == -1) errExit("lseek"); numRead = read(fd, buf, length); if (numRead == -1) errExit("read"); printf("Read %ld bytes\n", (long) numRead); exit(EXIT_SUCCESS); } 

是想展示以下三点：

• 用于传递数据的缓冲区，其内存边界必须对其为块大小的整数倍。
• 数据传输的开始的点，亦即文件和设备的偏移量，必须是块大小的整数倍。
• 待传递数据的长度必须是块大小的整数倍。

中间大块的注释里有一句话：

we ensure that 'buf' is aligned on a non-power-of-two multiple of 'alignment'.

请问这个用意是什么呢？（因为我感觉这是不必要的，为什么非得让 buf 的开始地址对其为非 2 的幂呢？）