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C語言大俠們，如何詳細解讀這兩個函數(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;
}

//b24e是一個返回char類型指針的函數(shù)，函數(shù)帶3個參數(shù)。一個signed char類型指針，一個unsigned char類型指針，一個堆棧塊數(shù)量的參數(shù)
char *b24e(char *buf, unsigned char *byst, size_t sizeOfBytes)
{
    int i = 0;
    
    //指針p指向形參(*byst)
    unsigned char *p = byst;
    
    //第一次循環(huán)(i = p - byst) = 0強制轉換成(size_t)類型跟參數(shù)sizeOfBtes比較。結果大概率是真
    while ((size_t)(i = (p-byst)) < sizeOfBytes) {
        
        //sel是數(shù)組。數(shù)據(*p)右移4位作為sel數(shù)組的標號。(sel[((*p) >> 4)]是作32位數(shù)據處理，并且是在做類似壓棧操作。buf[2 * i]是char類型數(shù)組，搞不懂為什么這樣做？)
        buf[2*i] = sel[((*p) >> 4)];
        
        //((*p) & 0x0f)是上面右移了的4位，最小是0，最大是15。sel[8~23]中的一個字節(jié)數(shù)據賦值給buf[(2*i)+1]
        buf[(2*i)+1] = sel[23 - ((*p) & 0x0f)];
      
       //指針加一
        p++;
    }
    
    //數(shù)組byst處理完后在buf后面加0，數(shù)組bug[]現(xiàn)在存的是字符串數(shù)據
    buf[(2*i)+1] = '\0';
    
   //返回指針buf
    return buf;
}

太難了，我也不知道什么意思，歡迎大神指正錯誤。