求救：gnu strncpy 的实现代码看不懂.
gnu 中 strncpy 的实现代码看不懂.，大家帮解释一下，让我理解它：

C/C++ code

 
 char* __strncpy(char* dest, const char* src, size_t n) 
 { 
 char c; 
 char *s = dest; 
 if (n >= 4) 
     { 
 size_t n4 = n >> 2; 
 for (;;) 
 { 
 c = *src++; 
 *dest++ = c; 
 if (c == '\0') 
 break; 
 c = *src++; 
 *dest++ = c; 
 if (c == '\0') 
 break; 
 c = *src++; 
 *dest++ = c; 
 if (c == '\0') 
 break; 
 c = *src++; 
 *dest++ = c; 
 if (c == '\0') 
 break; 
 if (--n4 == 0) 
 goto last_chars; 
 } 
 n -= dest - s; 
 goto zero_fill; 
     } 
 last_chars: 
 n &= 3; 
 if (n == 0) 
 return dest; 
 for (;;) 
     { 
 c = *src++; 
 --n; 
 *dest++ = c; 
 if (c == '\0') 
 break; 
 if (n == 0) 
 return dest; 
     } 
 zero_fill: 
 while (n-- > 0) 
 dest[n] = '\0'; 
 return dest - 1; 
 } 
 

C/C++ code

char* __strncpy(char* dest, const char* src, size_t n){    char c;    char *s = dest;    if (n >= 4) //如果大于4的话，即 二进制大于 100    {        size_t n4 = n >> 2; //右移两位跟大于4联系发现这样移位后能大于0        for (;;)//死循环一次拷贝四个字节，如果遇到0跳出，将剩余空间填0，否则每次拷贝4个字节直到拷贝完4的整数倍个字符        {            c = *src++;            *dest++ = c;            if (c == '\0')                break;            c = *src++;            *dest++ = c;            if (c == '\0')                break;            c = *src++;            *dest++ = c;            if (c == '\0')                break;            c = *src++;            *dest++ = c;            if (c == '\0')                break;            if (--n4 == 0)                goto last_chars;// 不足4的整数倍        }        n -= dest - s;// 拷贝了多少个字符        goto zero_fill;//遇到了0，于是剩余均填0    }last_chars://不足4的部分    n &= 3;//还有字符没？    if (n == 0)//没了返回        return dest;        for (;;)//有的话继续拷贝，结束填0    {        c = *src++;        --n;        *dest++ = c;        if (c == '\0')            break;        if (n == 0)            return dest;    }    zero_fill://遇到o后，全部填零    while (n-- > 0)        dest[n] = '\0';        return dest - 1;}
[解决办法]
一次拷贝4字节自然比一次拷贝1字节快.