std::_throw_out_of_range 不知从何而来

Question

我是 C++ 的绝对初学者。字面上地。刚刚过去一周了。今天我正在编写一个程序来测试需要多少次迭代才能使某个数字成为回文。这是代码：

#include <iostream>
#include <string>
#include <algorithm>

/*  This program calculates the steps needed
    to make a certain number palindromic.
    It is designed to output the values for
    numbers 1 to 1000
*/
using namespace std;

class number
{
public:
    string value;
    void reverse();
};

void number::reverse()
{
    std::reverse(value.begin(),value.end());
}

void palindrome(number num)
{
    string n=num.value;
    number reversenum, numsum, numsumreverse;
    reversenum=num;
    reversenum.reverse();
    numsum.value=num.value;
    numsumreverse.value=numsum.value;
    numsumreverse.reverse();
    int i=0;
    while (numsum.value.compare(numsumreverse.value) !=0)
    {
        reversenum=num;
        reversenum.reverse();
        numsum.value=to_string(stoll(num.value,0,10)+stoll(reversenum.value,0,10));
        numsumreverse.value=numsum.value;
        numsumreverse.reverse();
        num.value=numsum.value;
        i++;
    }
    cout << "The number " << n << " becomes palindromic after " << i << " steps : " << num.value << endl;
}

int main()
{
    number temp;
    int i;
    for (i=1; i<1001; i++)
    {
        temp.value=to_string(i);
        palindrome(temp);
    }
    return 0;
}

对于 195 以内的数字，它会顺利进行。但是，在 196 的情况下，我会收到错误。它说：

抛出“std::out_of_range”实例后调用的终止what(): stoll

我不知道该怎么办。我尝试从 开始，196但错误仍然存​​在。任何帮助将不胜感激。:)

更新：这次我尝试使用 ttmath 库来做到这一点。但是啊！它再次停在 195，甚至没有报告错误！我可能正在做一些愚蠢的事情。任何意见将不胜感激。这是更新后的代码：

#include <iostream>
#include <string>
#include <algorithm>
#include <ttmath/ttmath.h>

/*  This program calculates the steps needed
    to make a certain number palindromic.
    It is designed to output the values for
    numbers 1 to 1000
*/
using namespace std;

class number
{
public:
    string value;
    void reverse();
};

void number::reverse()
{
    std::reverse(value.begin(),value.end());
}

template <typename NumTy>
string String(const NumTy& Num)
{
    stringstream StrStream;
    StrStream << Num;
    return (StrStream.str());
}

void palindrome(number num)
{
    string n=num.value;
    number reversenum, numsum, numsumreverse;
    reversenum=num;
    reversenum.reverse();
    numsum.value=num.value;
    numsumreverse.value=numsum.value;
    numsumreverse.reverse();
    ttmath::UInt<100> tempsum, numint, reversenumint;
    int i=0;
    while (numsum.value.compare(numsumreverse.value) !=0)
    {
        reversenum=num;
        reversenum.reverse();
        numint=num.value;
        reversenumint=reversenum.value;
        tempsum=numint+reversenumint;
        numsum.value=String<ttmath::UInt<100> >(tempsum);
        numsumreverse.value=numsum.value;
        numsumreverse.reverse();
        num.value=numsum.value;
        i++;
    }
    cout << "The number " << n << " becomes palindromic after " << i << " steps : " << num.value << endl;
}

int main()
{
    number temp;
    int i;
    for (i=196; i<1001; i++)
    {
        temp.value=to_string(i);
        palindrome(temp);
    }
    return 0;
}

更新：已经解决了。一些研究表明 196 可能是一个利克雷尔数。我在暗示 ttmath 库后得到的结果只是让我确信我的算法有效。我已经尝试过 10000 以内的所有数字，并且给出了完美的结果。这是最终的代码：

#include <iostream>
#include <string>
#include <algorithm>
#include <ttmath/ttmath.h>
#include <limits>

/*  This program calculates the steps needed
    to make a certain number palindromic.
    It is designed to output the values for
    numbers inside a desired range
*/
using namespace std;
string LychrelList;
int LychrelCount=0;

class number
{
public:
    string value;
    void reverse();
};

void number::reverse()
{
    std::reverse(value.begin(),value.end());
}

template <typename NumTy>
string String(const NumTy& Num)
{
    stringstream StrStream;
    StrStream << Num;
    return (StrStream.str());
}

void palindrome(number num)
{
    string n=num.value;
    number reversenum, numsum, numsumreverse;
    reversenum=num;
    reversenum.reverse();
    numsum.value=num.value;
    numsumreverse.value=numsum.value;
    numsumreverse.reverse();
    ttmath::UInt<100> tempsum, numint, reversenumint;
    int i=0;
    while ((numsum.value.compare(numsumreverse.value) !=0) && i<200)
    {
        reversenum=num;
        reversenum.reverse();
        numint=num.value;
        reversenumint=reversenum.value;
        tempsum=numint+reversenumint;
        numsum.value=String<ttmath::UInt<100> >(tempsum);
        numsumreverse.value=numsum.value;
        numsumreverse.reverse();
        num.value=numsum.value;
        i++;
    }
    if (i<200) cout << "The number " << n << " becomes palindromic after " << i << " steps : " << num.value << endl;
    else
    {
        cout << "A solution for " << n << " could not be found!!!" << endl;
        LychrelList=LychrelList+n+" ";
        LychrelCount++;
    }
}

int main()
{
    cout << "From where to start?" << endl << ">";
    int lbd,ubd;
    cin >> lbd;
    cout << endl << "And where to stop?" << endl <<">";
    cin >> ubd;
    cout << endl;
    number temp;
    int i;
    for (i=lbd; i<=ubd; i++)
    {
        temp.value=to_string(i);
        palindrome(temp);
    }
    if (LychrelList.compare("") !=0) cout << "The possible Lychrel numbers found in the range are:" << endl << LychrelList << endl << "Total - " << LychrelCount;
    cout << endl << endl << "Press ENTER to end the program...";
    cin.ignore(numeric_limits<streamsize>::max(), '\n');
    string s;
    getline(cin,s);
    cout << "Thanks for using!";
    return 0;
}

这是一个非常棒的社区。特别感谢Marco A . :)

再次更新：我设计了自己的 add() 函数，可以减少程序对外部库的依赖。它导致可执行文件更小，性能也更快。这是代码：

#include <iostream>
#include <string>
#include <algorithm>
#include <limits>

/*  This program calculates the steps needed
    to make a certain number palindromic.
    It is designed to output the values for
    numbers inside a desired range
*/
using namespace std;
string LychrelList;
int LychrelCount=0;

string add(string sA, string sB)
{
    int iTemp=0;
    string sAns;
    int k=sA.length()-sB.length();
    int i;
    if (k>0){for (i=0;i<k;i++) {sB="0"+sB;}}
    if (k<0) {for (i=0;i<-k;i++) {sA="0"+sA;}}
    for (i=sA.length()-1;i>=0;i--)
    {
        iTemp+=sA[i]+sB[i]-96;
        if (iTemp>9)
        {
            sAns=to_string(iTemp%10)+sAns;
            iTemp/=10;
        }
        else
        {
            sAns=to_string(iTemp)+sAns;
            iTemp=0;
        }
    }
    if (iTemp>0) {sAns=to_string(iTemp)+sAns;}
    return sAns;
}

void palindrome(string num)
{
    string n=num;
    string reversenum, numsum, numsumreverse;
    numsum=num;
    numsumreverse=numsum;
    reverse(numsumreverse.begin(),numsumreverse.end());
    int i=0;
    while ((numsum.compare(numsumreverse) !=0) && i<200)
    {
        reversenum=num;
        reverse(reversenum.begin(),reversenum.end());
        numsum=add(num,reversenum);
        numsumreverse=numsum;
        reverse(numsumreverse.begin(),numsumreverse.end());
        num=numsum;
        i++;
    }
    if (i<200) cout << "The number " << n << " becomes palindromic after " << i << " steps : " << num << endl;
    else
    {
        cout << "A solution for " << n << " could not be found!!!" << endl;
        LychrelList=LychrelList+n+" ";
        LychrelCount++;
    }
}

int main()
{
    cout << "From where to start?" << endl << ">";
    int lbd,ubd;
    cin >> lbd;
    cout << endl << "And where to stop?" << endl <<">";
    cin >> ubd;
    cout << endl;
    string temp;
    int i;
    for (i=lbd; i<=ubd; i++)
    {
        temp=to_string(i);
        palindrome(temp);
    }
    if (LychrelList.compare("") !=0) cout << "The possible Lychrel numbers found in the range are:" << endl << LychrelList << endl << "Total - " << LychrelCount;
    cout << endl << endl << "Press ENTER to end the program...";
    cin.ignore(numeric_limits<streamsize>::max(), '\n');
    string s;
    getline(cin,s);
    cout <<endl << "Thanks for using!";
    return 0;
}

你们在这里给了我很多帮助，让我找到了自己的路。感谢大家。:)

Answer 1

你溢出了， 因为和 的long long最后两个有效值是 7197630720180367016 和 6107630810270367917，它们加在一起，远远高于 a 的最大大小（我的机器上是 9223372036854775807 ）。这将产生负值并破坏您的下一次调用num.valuereversenum.valuelong longstoll

如果转换后的值超出结果类型的范围或者底层函数（std::strtol 或 std::strtoll）将 errno 设置为 ERANGE，则抛出 std::out_of_range 。

（参考）

• 如果您试图获得下一个最小的回文，您应该使用另一种方法，就像我在这里解释的那样。

你可以Live Example在这里找到一个

• 如果您更喜欢/必须继续您的方法，您应该在字符串上手动添加或使用bigint库（再次查看此处并根据plusOne()您的喜好修改函数）