在向量中访问指针的成员函数时的分段故障 [英] Segmentation fault when accessing a pointer's member function in a vector

问题描述

由于某种原因，我得到一个分段错误;我使用一个指针的向量，它指向一个类对象。基本上我需要一个具有指向其他节点的指针向量的节点，在其他节点上做一个多图。以下是我的代码的相关部分：

For some reason I'm getting a segmentation fault; I'm using a vector of pointers, which point to a class object. Basically I need a node that has a vector of pointers to other nodes, in other to make a multigraph. Here's the relevant part of my code:

node.h：

#ifndef NODE_H
#define NODE_H

class node
{
public:
    string content()
    vector<node*> next;      //causing the error
    void add_arc(node a);

    string rna_frag;

#endif

node.cpp：

void node::add_arc(node a)
{
    node *b = &a;    //b->content() works fine here
    next.push_back(b);
}

string node::content()
{
    return rna_frag;
}

main.cpp： b
$ b

main.cpp:

int main()
{
    vector<node> nodes;
    node a;
    node b;
    node c;

    a.add_arc(b);
    a.add_arc(c);
    a.rna_string = "G";

    nodes.push_back(a);
    nodes.push_back(b);
    nodes.push_back(c);

    cout << nodes[0].content() << endl;    //prints "G", works fine
    cout << nodes[0].next.size() << endl;  // prints "2", works fine
    cout << nodes[0].next[0]->content() << endl;  //segmentation fault
    //cout << nodes[0].next->content() << endl;   //also segmentation fault
    //cout << nodes[0].next[0]->rna_frag << endl; //also segmentation fault
}

在这种情况下，nodes [0]是G，并指向其他2个节点，所以前2个couts工作完美。但是当我访问的矢量的内容，它只是崩溃，并给出一个分割错误错误。任何人都知道为什么？

in this case, nodes[0]'s string is "G" and is pointing to 2 other nodes, so the first 2 couts are working perfectly. But when I access the vector's contents it just crashes and gives a segmentation fault error. Anyone know why?

推荐答案

在 add_arc 参数 a ，然后在函数退出时被销毁 - 因此你有未定义的行为。

In add_arc you are storing the address of the parameter a, which is then destroyed when the function exits - so you have undefined behaviour.

当你调用 nodes.push_back（）时，也会复制节点，这将导致你很多的悲伤。

You're also copying nodes when you call nodes.push_back(), which is going to cause you a lot of grief.

你将需要停止复制或写一个正确的复制构造函数（然后遵循3或5的规则）。

You are going to need to either stop copying or write a proper copy constructor (and then follow the rule of 3, or 5).

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