比较C中的两个int数组

Is there a better way to compare two int array in C++.

有更好的方法比较c++中的两个int数组吗?

struct integer
{
    int *number;
    int size;
};

This is my struct and i have used this comparing function. I found this solution but it looks like bulshit.

这是我的结构，我用了这个比较函数。我找到了这个解，但它看起来像扯淡。

int BigInteger::biggest(integer number1, integer number2)
{
    int biggest = 0;
    if (number1.size > number2.size)
    {
        biggest = 1;
    }
    else if (number1.size < number2.size)
    {
        biggest = -1;
    }
    else
    {
        for (int i = number1.size; i >=0 ; --i)
        {
            if (number1.number[i] > number2.number[i])
            {
                biggest = 1;
                return biggest;
            }
            else if (number1.number[i] < number2.number[i])
            {
                biggest = -1;
                return biggest;
            }
            else
            {
                continue;
            }
        }
    }
    return biggest;
}

if function turn value as 1 then number1 greater than number2
if function turn value as 0 then number1 equal to number2
if function turn value as -1 then number2 greater than number1

如果函数的值是1，那么数字1大于数字2如果函数的值是0，那么数字1等于数字2如果函数的值是-1，那么数字2大于数字1

3 个解决方案

#1

This will not improve anything except, that it's much clearer

这将不会改善任何事情，除非，它更加清晰。

int BigInteger::biggest(integer number1, integer number2)
{
    if (number1.size != number2.size)
        return number1.size - number2.size;
    for (int i = 0 ; i < number1.size; ++i) {
        if (number1.number[i] == number2.number[i])
            continue;
        return number1.number[i] - number2.number[i];
    }
    return 0;
}

This function

这个函数

Returns > 0 then number1 > number2
返回>，然后是1 > 2
Returns < 0 then number1 < number2
返回< 0，然后返回number1 < number2
And 0 if they are equal
如果它们是相等的。

Also, in c++ you can overload operators so implementing > and < should be easy using this code.

此外，在c++中，可以重载操作符，因此使用此代码实现>和 <应该很容易。< p>

#2

There are a few small bugs. You don't need some of this code. It will run slightly faster when it is not part of a class. And with just a bit of modification it will work with qsort() (the parameters need to be pointers) (edit: void pointers)

有一些小bug。你不需要这些代码。当它不是类的一部分时，它会运行得稍微快一点。只要稍加修改就可以使用qsort()(参数必须是指针)(编辑:void指针)

struct integer
{
    int *number;
    int size;
}; 

int biggest(const integer &number1, const integer &number2)
{
    int biggest = 0;
    if (number1.size > number2.size)
    {
        biggest = 1;
    }
    else if (number1.size < number2.size)
    {
        biggest = -1;
    }
    else
    {
        for (int i = number1.size - 1; i >= 0; i--) // or --i, no difference
        {
            if (number1.number[i] > number2.number[i])
            {
                biggest = 1;
                return biggest;
            }
            if (number1.number[i] < number2.number[i])
            {
                biggest = -1;
                return biggest;
            }
            // don't need a continue statement here
        }
    }
    return biggest;
}

#3

You can use the std::mismatch() function like this:

您可以使用std:失配()函数如下:

int BigInteger::biggest(integer a, integer b)
{
    if (a.size != b.size)
        return a.size - a.size;
    const auto res = std::mismatch(a.number, a.number + a.size, b.number);
    if (res.first == (a.number + a.size))
        return 0;
    else
        return (*res.first) < (*res.second) ? -1 : 1;
}

The name of the function biggest() is quite misleading. compare() would be much better. Also, if the function's return type would be boolean (true: a < b), then you could use std::lexicographical_compare(), and the code would much more simpler.

最大()函数的名称很容易引起误解。compare()会更好。此外，如果函数的返回类型是布尔型(true: a < b)，那么可以使用std:: lexicographicoleical_compare()，代码会简单得多。

#1