c++中大数类的递归

c++ recursion in large number class

本文关键字:递归 c++      更新时间:2023-10-16

我正在实现一个大量的库,并且我被一个奇怪的问题卡住了。我有一个乘法的递归实现,但是第一个递归调用甚至没有开始解析。下面是代码的一部分(这是在BigInt类中):

BigInt multiply_utility(BigInt num1, BigInt num2)
{
    cout << "Location 1";
    if(num1.get_length() == 1)
    {
        if(num1.get(MAX-1) == 0 || num2.get(MAX-1) == 0)
            {
                BigInt zero;
                return zero;
            }
        BigInt temp = num2;
        for(int i = 1; i < num1.get(MAX-1); i++)
            num2 = temp.add(num2);
        return num2;
    }
    else if(num2.get_length() == 1)
    {
        if(num1.is_zero() || num2.is_zero())
        {
            BigInt zero;
            return zero;
        }
        BigInt temp = num1;
        for(int i = 1; i < num2.get(MAX-1); i++)
            num1 = temp.add(num1);
        return num1;
    }
    int m = max(num1.get_length(), num2.get_length());
    BigInt low1, low2, high1, high2;

    for(int i = MAX-num1.get_length(); i < MAX-(num1.get_length()/2); i++)
        low1.set((MAX-((MAX-(num1.get_length()/2))-i)) ,num1.get(i));
    low1.auto_set_length();

    for(int i = MAX-num2.get_length(); i < MAX-(num2.get_length()/2); i++)
        low2.set((MAX-((MAX-(num2.get_length()/2))-i)) ,num2.get(i));
    low2.auto_set_length();
    for(int i = MAX-(num1.get_length()/2); i < MAX; i++)
        high1.set(i,num1.get(i));
    high1.auto_set_length();
    for(int i = MAX-(num2.get_length()/2); i < MAX; i++)
        high2.set(i,num2.get(i));
    high2.auto_set_length();
    cout << "low1 = " << low1.str() << endl;
    cout << "high1 = " << high1.str() << endl;
    cout << "low2 = " << low2.str() << endl;
    cout << "high2 = " << high2.str() << endl;
    BigInt z0,z1,z2;
    BigInt handle;
    cout << "Location 2";
    z0 = multiply_utility(low1,low2);
    cout << "Location 3";
    z1 = multiply_utility(low1.add(high1),low2.add(high2));
    z2 = multiply_utility(high1,high2);

    BigInt a;
    a = z1.subtract(z2);
    a = a.subtract(z0);
    return (((z2.shift(m)).add(a.shift(m/2))).add(z0));
    /*
    return (z2*10^(m))+((z1-z2-z0)*10^(m/2))+(z0)
    */
}

它先打印"location 1",然后再打印"location 2",但是不像递归调用那样再次打印"location 1"。你知道哪里出错了吗?它可能是一个类方法函数吗?

你不能在那里放一个断点来调试它吗?

我怀疑代码在某个时候击中了return并弹出了这个函数,因此,没有像您期望的那样显示"location…"。

好了,各位,这似乎是一个堆栈溢出。我创建的对象太大,无法全部放入堆栈中,而递归又使堆栈负担过重。谢谢你无论如何!