C語(yǔ)言大俠們，如何詳細(xì)解讀這兩個(gè)函數(shù)？？？


char *b24e(char *buf, unsigned char *byst, size_t sizeOfBytes)
{
    int i = 0;
    unsigned char *p = byst;
    while ((size_t)(i = (p-byst)) < sizeOfBytes) {
        buf[2*i] = sel[((*p) >> 4)];
        buf[(2*i)+1] = sel[23 - ((*p) & 0x0f)];
        p++;
    }
    buf[(2*i)+1] = '\0';
    return buf;
}

unsigned char *b24d(unsigned char *buf, char *str, size_t countOfChars)
{
    size_t i;
    char *p = str;
    char *loc[2];
    unsigned char n[2];
    if (countOfChars % 2)
        return NULL;
    for (i = 0; i < (countOfChars>>1); i++) {

        loc[0] = strchr( sel, str[2*i] );
        loc[1] = strchr( sel, str[ ( 2*i ) + 1 ] );
        if (loc[0] == NULL || loc[1] == NULL)
            return NULL;
        n[0] = (unsigned char)( loc[0] - sel );
        n[1] = 23 - (unsigned char)( loc[1] - sel );
        buf = (unsigned char)((n[0] << 4) | n[1]);
    }
    return buf;
}