Question

#include <stdio.h>
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <fstream>
#include <utility>
using namespace std;

struct node
{
    int level = -1;
    int value = -5;
    node *left;
    node *right;
};

int array[100][100];
void storetree(node *root, int val);
void printtree();

int main()
{
    cout << " u there?" << endl;

    for (int i = 0; i < 100; i++)
    {
        for (int j = 0; j < 100; j++)
        {
            array[i][j] = -5;
        }
    }

    ifstream file1;
    ifstream file2;
    file1.open("tree1.txt");
    file2.open("graph1.txt");
    node root;
    node *root1;
    int val1;
    int randval;

    file1 >> val1;
    root.level = 0;
    root1->level = 0;
    root.value = val1;
    root1->value = val1;
    array[0][0] = val1;

    cout << "made it" << endl;

    root1->left->value = -5;     // <-- Error happens here
    root1->right->value = -5;
    root1->left->level = -5;
    root1->right->level = -5;

So my error happens when I access root1->left->value, and I get the stack dump error.

Is it impossible to access root1->left->value the way I've written it? Through print statements I've deduced that this is the error. I don't understand pointers well and would appreciate help. :)

Answer 1

我在下面注释了您的源代码：

...

// Allocate an actual instance of your 'node' struct (local variable, on the stack)
node root;

// This allocates a POINTER variable, which just contains an address.
// The compiler knows that 'root1' will point at 'node' types.
// Note that you've only created a pointer variable here; NOT an actual node.
// Further, you haven't initialized it, so its value is "undefined"
// (you can't assume you know what its value is ... it could be anything).
node *root1;

...

// Set 'level' on the root node to zero
root.level = 0;    // This is OK, because root is an actual 'node' struct instance

// Set 'level' ON THE NODE THAT 'root1' POINTS TO to zero
root1->level = 0;  // This is not OK, because root1 is a wild (uninitialized) pointer

这是你的第一个问题。你创建了一个指针，但你没有把它指向任何东西。 root1（它引用的地址）的值是 undefined （可能是NULL，可能是该变量所在的内存中的任何内容）

首先，最好在定义局部变量时将其初始化。未初始化的变量可能具有未定义的值，并且可以添加无数小时调试你的生活，你的程序的每次运行都会做与上一次不同的事情。的Bleh。

如果在定义变量时尚未准备好分配实际值，将它设置为某些已知的值，如0，nullpointer，等等。这样，如果你忘记稍后再设置它，至少你的程序每次都会做错误的事情。

node *root1 = nullptr; // (...or maybe your compiler uses "NULL" or "(void *)0"...)

看起来您将根据从输入文件中读取的内容构建节点树？如果是这样，那么你几乎肯定会动态地分配节点结构，因为你无法提前知道你需要多少。

// Allocate and initialize a new node struct (on the heap) root1 = new node(); // Set 'level' ON THE NODE THAT 'root1' POINTS TO to zero root1->level = 0; // Yay, now this works

变量'root1'现在包含新分配的'node'结构的地址。其余代码应该可以在那里工作。

只是提醒一下，'正确'程序（例如，不泄漏内存的程序）应该最终调用delete来调用new所返回的每个指针。在构建动态分配的node对象树时，请记住这一点;当你完成所有操作时，你需要在每个上面调用delete（除了root，你没有动态分配）。

2 [main] hw3 10368 cygwin_exception::open_stackdumpfile: Dumping stack trace to hw3.exe.stackdump

1 个答案: