Question

我正在为Project Euler开发一个程序来添加2 ^ 1000的所有数字。到目前为止，我已经能够跟踪程序分段错误，当它达到大约5位数并试图将一个数据推到函数carry（）中第61行的向量上。

#include <iostream>
#include <vector>
#include <string>
using namespace std;

class MegaNumber
{
        vector<int> data; //carries an array of numbers under ten, would be char but for simplicity's sake
        void multiplyAssign(int operand, int index); //the recursive function called by the *= operator
        void carry(int index);//if one of the data entries becomes more than ten call this function
    public:
        void printNumber(); //does what it says on the can
        void operator*=(MegaNumber operand);
        void operator*=(int operand);
        void operator+=(int operand);
        MegaNumber(string);
        unsigned long int AddAllDigits();//returns the value of all of the digits summed
};

MegaNumber::MegaNumber(string operand)
{
    for(int i= operand.size()-1; i>=0;i--) //run it into the memory smallest digit first
    {
        data.push_back(operand[i]-48); //converts a text char to an int
    }
}

void MegaNumber::printNumber()
{
    int temp = data.size();
    for(unsigned int i=(temp); i>0;--i)
    {
     cout << (int)data[i-1];
    }
}

void MegaNumber::operator*=(int operand)
{
    if(operand > 9)
    {
        cout << "function does not yet deal with large ints than 9";
    }
    else multiplyAssign(operand, 0);
}

void MegaNumber::multiplyAssign(int operand, int index)
{
    data[index] *=operand;
    if(index<data.size()) multiplyAssign(operand, index+1);
    if(data[index] > 9) carry(index);
}

void MegaNumber::carry(int index)
{

    int temp = (data[index] / 10); //calculate the amount to carry
    if(data.size()==index+1)
    {
     data.push_back(temp);//if there is no upper digit push it onto the stack
    }
    else
    {
        data[index+1]+=temp; //else add it to the next digit

        if(data[index+1]>9) carry(index+1); //rinse and repeat
    }
     data[index]-=temp*10; //remove what's been carried
}

unsigned long int MegaNumber::AddAllDigits() //does what it says on the can
{
    unsigned long int Dagger = 0;
    for(int i=0; i<data.size();i++) Dagger+=data[i];
    return Dagger;
}

int main()
{
    MegaNumber A("2");
    A.printNumber();
    cout << "\n";
    for(unsigned int i=0; i<20; i++) A*=2;
    A.printNumber();
    cout << "\n";
    cout << A.AddAllDigits() << "\n";
    cout << "Hello world!" << endl;
    return 0;
}

可能导致这种情况的原因是什么？

Answer 1

在multiplyAssign之前检查它是否是有效索引之前使用数据[index]：

data[index] *= operand;
if(index<data.size()) multiplyAssign(operand, index+1);

同时使用'0'代替48。这更容易，更清晰，并且不易出错。

Answer 2

void MegaNumber::multiplyAssign(int operand, int index)
{
    data[index] *=operand;
    if(index<data.size()) multiplyAssign(operand, index+1);
    if(data[index] > 9) carry(index);
}

index基于0，而data.size()基于1，所以说，data.size()返回的数字1大于最大有效index。所以看起来你的意图是

if( index < data.size() - 1) multiplyAssign(operand, index+1);

然后它起作用。
附：将代码分成几行，任何必须维护代码的人都会感谢你：

if (index < data.size() - 1) 
{
    multiplyAssign(operand, index + 1);
}

Answer 3

我认为问题可能在这里：data[index+1]+=temp;

如果index参数eq，则该元素不可能存在。大小为data。

所以，我的建议：

使用迭代器访问std::vector
如果您不使用Iterators，请检查绑定条件

导致vector <int>中的push_back导致分段错误似乎是简单的操作</int>

3 个答案: