c++ - 简单的运算符(operator)+问题躲避我

Question

我试图添加两个 uint8_t 列表，就好像列表是单独的整数一样，我得到了一些奇怪的值:

0x1000 + 0x100 + 0x10 -> 0x1210 ??????

代码如下:

// values 0x123456 stored as: {12, 34, 56}
integer operator+(integer rhs){
    // internal list called 'value'
    std::list <uint8_t> top = value, bottom = rhs.value;
    if (value.size() < rhs.value.size())
        top.swap(bottom);
    top.push_front(0);                           // extra byte for carrying over
    while (bottom.size() + 1 < top.size())       // match up the byte sizes, other than the carry over
        bottom.push_front(0);
    bool carry = false, next_carry = false;
    for(std::list <uint8_t>::reverse_iterator i = top.rbegin(), j = bottom.rbegin(); j != bottom.rend(); i++, j++){
        next_carry = (((uint8_t) (*i + *j + carry)) <= std::min(*i, *j));
        *i += *j + carry;
        carry = next_carry;
    }
    if (carry)
        *top.begin() = 1;
return integer(top);
}

谁能告诉我我做错了什么？

Answer 1

在您的示例 ( 0x100 + 0x10 ) 中，您以 carry = false 开头, *top.rbegin() = 0 , 和 *bottom.rbegin() = 0 .当我们进入循环时，我们会看到以下测试:

next_carry = (((uint8_t) (*i + *j + carry)) <= std::min(*i, *j));
// given *i == 0, *j == 0, and carry == false
// this will evaluate to TRUE

自 next_carry滚动到下一个添加，您最终得到 carry = true , 当它应该是 false . 将条件切换为 < std::min(*i, *j) .