glibc源码解读——memset

#include <string.h>#include <memcopy.h>#undef memsetvoid *inhibit_loop_to_libcallmemset(void *dstpp, int c, size_t len){    long int dstp = (long int)dstpp;    if (len >= 8)    {        size_t xlen;        op_t cccc;        cccc = (unsigned char)c;        cccc |= cccc << 8;        cccc |= cccc << 16;        if (OPSIZ > 4)            /* Do the shift in two steps to avoid warning if long has 32 bits.  */            cccc |= (cccc << 16) << 16;        /* There are at least some bytes to set.        No need to test for LEN == 0 in this alignment loop.  */        while (dstp % OPSIZ != 0)        {            ((byte *)dstp)[0] = c;            dstp += 1;            len -= 1;        }        /* Write 8 `op_t' per iteration until less than 8 `op_t' remain.  */        xlen = len / (OPSIZ * 8);        while (xlen > 0)        {            ((op_t *)dstp)[0] = cccc;            ((op_t *)dstp)[1] = cccc;            ((op_t *)dstp)[2] = cccc;            ((op_t *)dstp)[3] = cccc;            ((op_t *)dstp)[4] = cccc;            ((op_t *)dstp)[5] = cccc;            ((op_t *)dstp)[6] = cccc;            ((op_t *)dstp)[7] = cccc;            dstp += 8 * OPSIZ;            xlen -= 1;        }        len %= OPSIZ * 8;        /* Write 1 `op_t' per iteration until less than OPSIZ bytes remain.  */        xlen = len / OPSIZ;        while (xlen > 0)        {            ((op_t *)dstp)[0] = cccc;            dstp += OPSIZ;            xlen -= 1;        }        len %= OPSIZ;    }    /* Write the last few bytes.  */    while (len > 0)    {        ((byte *)dstp)[0] = c;        dstp += 1;        len -= 1;    }    return dstpp;}libc_hidden_builtin_def(memset)

其中部分定义

#define op_t unsigned long int#define OPSIZ(sizeof(op_t))typedef unsigned char byte;

考虑到效率的因素：

1、len < 8 的时候，按每次一个字节来读写。

2、len >= 8 的时候，32位机器按每次 4 个字节来读写，64位机器按每次 8 个字节来读写，连续做 8 次。剩下的接着按 每次 4 或 8 字节读写， 每次 1 字节读写。

疑问：

假如在32位机器下，每次取得 dstp 的地址都是 4 的倍数吗？怎么才能取得 不是 4 倍数的地址？