分段故障核心转储向量指针

Question

void printTop5()
{
    int counter = 1;
    int x, y;
    float civ;
    cout << "Total no. of records available = " << pointVector.size();
    cout << "Printing top 5 exploration destinations....";
    sort(pointVector.begin(), pointVector.end(), ascendingCiv);
    for (int i = 0; i < 5; i++)
    {
        PointTwoD p1 = pointVector[i];
        x = p1.getX();
        y = p1.getY();
        civ = p1.getCivIndex();
        cout << counter << ")\t";
        if (civ > 0)
        {
            cout << "Civ idx: " << civ << " ; at sector(" << x << "," << y < ")\n";
        }
        else
            cout << "<No records available>";
    }
    }

bool ascendingCiv(const PointTwoD &d1, const PointTwoD &d2)
{
    return d1.getCivIndex() > d2.getCivIndex();
}

当我运行这个函数时，我得到这个错误，它说分段错误核心转储。任何的想法？听说是内存问题。

Answer 1

一个问题是您没有检查向量中是否确实有至少 5 个元素：

for (int i = 0; i < 5; i++)
{
    PointTwoD p1 = pointVector[i];

如果 i 越界，访问 pointVector[i] 将调用未定义的行为。

循环条件可以改成如下：

#include <algorithm>
//...
size_t loopEnd = std::min(pointVector.size(), 5);
for (size_t i = 0; i < loopEnd; i++)
{
    PointTwoD p1 = pointVector[i];

循环将达到向量的大小或 5 个元素，以较小者为准。

分段故障核心转储向量指针

Segmentation fault core dumped vector pointer

c++

pointers

segmentation-fault

fault