当我输入12位数时，为什么以下代码会崩溃？ [重复]

Question

这个问题在这里已有答案：

• cin >> fails with bigger numbers but works with smaller ones? 4个答案

我一直在关注Coursera上的算法课程，我试着把我学到的东西放到代码中。这应该是一个“分而治之”的算法，我希望这部分是正常的。我有一个问题，我遇到它只是弄乱它：一切正常，直到我输入一个12位数的程序。当我这样做时，它只是结束cin并输出所有先前排序的数字（如果之前没有数字，则为空格）。如果可以的话，请告诉我如果发现错误有什么问题。这是我的代码：

#include "stdafx.h"
#include <iostream>
#include <vector>

using namespace std;

// setup global variable for the number of inversions needed
int inversions = 0;

// function to merge 2 sublists into 1 sorted list
vector<int> Merge_and_Count(vector<int>& split_lo, vector<int>& split_hi) {
    // setup output variable -> merged, sorted list of the 2 input sublists
    vector<int> out;
    int l = 0;
    int m = 0;

    // loop through all the elements of the 2 sublists
    for (size_t k = 0; k < split_lo.size() + split_hi.size(); k++) {
        // check if we reached the end of the first sublist
        if (l < split_lo.size()) {
            // check if we reached the end of the second sublist
            if (m < split_hi.size()) {
                // check which element is smaller and sort accordingly
                if (split_lo[l] < split_hi[m]) {
                    out.push_back(split_lo[l]);
                    l++;
                }
                else if (split_hi[m] < split_lo[l]) {
                    out.push_back(split_hi[m]);
                    m++;
                    inversions++;
                }
            }
            else {
                out.push_back(split_lo[l]);
                l++;
                inversions++;
            }
        }
        else {
            out.push_back(split_hi[m]);
            m++;
        }
    }

    return out;
}

// function that loops itself to split input into halves until it reaches the base case (1 element array)
vector<int> MergeSort_and_CountInversions(vector<int>& V) {
    // if we reached the base case, terminate the loop and feed the output to the previous loop to be processed
    if (V.size() == 1) return V;
    // if we didn't reach the base case
    else {
        // continue halving the sublists
        size_t const half_size = V.size() / 2;
        vector<int> split_lo(V.begin(), V.begin() + half_size);
        vector<int> split_hi(V.begin() + half_size, V.end());

        // feed them back into the loop
        return Merge_and_Count(MergeSort_and_CountInversions(split_lo), MergeSort_and_CountInversions(split_hi));
    }
}

// main function of the app, runs everything
int main()
{
    // setup main variables
    int input;
    vector<int> V;

    // get input
    cout << "Enter your numbers to be sorted (enter Y when you wish to proceed to the sorting)." << endl;
    cout << "Note: do NOT use duplicates (for example, do not input 1 and 1 again)!" << endl;
    while (cin >> input)
        V.push_back(input);

    cout << "\nThe numbers you chose were: " << endl;
    for (size_t i = 0; i < V.size(); i++)
        cout << V[i] << " ";

    // get sorted output
    vector<int> sorted = MergeSort_and_CountInversions(V);
    cout << "\n\nHere are your numbers sorted: " << endl;
    for (size_t j = 0; j < sorted.size(); j++)
        cout << sorted[j] << " ";

    // show number of inversions that were needed
    cout << "\n\nThe number of inversions needed were: " << inversions << endl;

    return 0;
}

Answer 1

12位十进制数字太长，无法容纳32位数字，这通常表示int。因此，使用>>读取该数字将失败，并且cin >> input将转换为false值，从而终止循环。

有关处理故障模式的详细信息，请参阅operator >> documentation。

Answer 2

您可以使用std::numeric_limits::digits10常量获取可由类型表示的最大基数为10的数字：

std::cout << std::numeric_limits<int>::digits10 << '\n';

有可能是int类型的最大有效位数为9，并且您尝试通过标准输入提供12。程序不会崩溃，(cin >> input)的条件只是评估为false。

Answer 3

对于32位整数，12位数太多，尝试使用unsigned long long int，检查这些限制：http://www.cplusplus.com/reference/climits/