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  • 摘:memcpy的实现

    摘:memcpy的实现

    https://baike.baidu.com/item/memcpy/659918?fr=aladdin

    void *memcpy(void *to, const void *from, size_t n)
    {
        void *xto = to;
        size_t temp, temp1;
    
        if (!n)
            return xto;
        if ((long)to & 1) {
            char *cto = to;
            const char *cfrom = from;
            *cto++ = *cfrom++;
            to = cto;
            from = cfrom;
            n--;
        }
        if (n > 2 && (long)to & 2) {
            short *sto = to;
            const short *sfrom = from;
            *sto++ = *sfrom++;
            to = sto;
            from = sfrom;
            n -= 2;
        }
        temp = n >> 2;
        if (temp) {
            long *lto = to;
            const long *lfrom = from;
    #if defined(CONFIG_M68000) || defined(CONFIG_COLDFIRE)
            for (; temp; temp--)
                *lto++ = *lfrom++;
    #else
            asm volatile (
                "    movel %2,%3
    "
                "    andw  #7,%3
    "
                "    lsrl  #3,%2
    "
                "    negw  %3
    "
                "    jmp   %%pc@(1f,%3:w:2)
    "
                "4:    movel %0@+,%1@+
    "
                "    movel %0@+,%1@+
    "
                "    movel %0@+,%1@+
    "
                "    movel %0@+,%1@+
    "
                "    movel %0@+,%1@+
    "
                "    movel %0@+,%1@+
    "
                "    movel %0@+,%1@+
    "
                "    movel %0@+,%1@+
    "
                "1:    dbra  %2,4b
    "
                "    clrw  %2
    "
                "    subql #1,%2
    "
                "    jpl   4b"
                : "=a" (lfrom), "=a" (lto), "=d" (temp), "=&d" (temp1)
                : "0" (lfrom), "1" (lto), "2" (temp));
    #endif
            to = lto;
            from = lfrom;
        }
        if (n & 2) {
            short *sto = to;
            const short *sfrom = from;
            *sto++ = *sfrom++;
            to = sto;
            from = sfrom;
        }
        if (n & 1) {
            char *cto = to;
            const char *cfrom = from;
            *cto = *cfrom;
        }
        return xto;
    }
    

    2018-11-08

      汇编部分没有看懂。

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  • 原文地址:https://www.cnblogs.com/igfirstblog/p/9929978.html
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