如何一次写一个单词给* char？

Question

我试图实现一个版本的memset来一次写一个字而不是逐个字节。 我目前使用的代码是：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

/* set in memory one word at a time
    dest:  destination 
    ch:    the character to write
    count: how many copies of ch to make in dest */
void wmemset(char *dest, int ch, size_t count) {
    long long int word;                    // 64-bit word
    unsigned char c = (unsigned char) ch;  // explicit conversion
    int i, loop, remain;

    for (i = 0, word = 0; i < 8 ; ++i) {
        word |= ((long long int)c << i * 8);
    }

    loop = count / 8;
    remain = count % 8;

    for (i = 0; i < loop; ++i, dest += 8) {
        *dest = word;    // not possible to set 64-bit to char
    }
    for (i = 0; i < remain; ++i, ++dest) {
        *dest = c;
    }
}

int main() {
    char *c;
    c = (char *) malloc(100);
    wmemset(c, 'b', 100);

    for (int i = 0; i < 100; i++)
        printf("%c", c[i]);

}

我认为64位字会溢出到指针的其余字节，但它看起来并非如此。 如何一次设置一个单词？

编辑：添加了一些评论并添加了主要功能。