c++ - cpp此功能的异常行为

我正在尝试编写一个计算间隙启发式的函数。下面是我的代码：

#include <iostream>
using namespace std;
#include <vector>
#include <map>
#include <string>
using namespace std;
int main(int argc, char const *argv[])
{
    string direction = "backward";
    string state_string = "012345";
    string goal_state_string = "125430";
    int n = 3;

    string ignored_pancakes;
    int gap = 0;
    state_string += to_string(state_string.length());
    unsigned int goal_state_index;
    goal_state_string += to_string(goal_state_string.length());
    if (direction == "forward")
    {
        for (unsigned int i = 0; i < n; i++)
        {
            ignored_pancakes += goal_state_string[i];
        }
    }
    else
    {
        for (unsigned int i = 0; i < n; i++)
        {
            ignored_pancakes += state_string[i];
        }

    }

    for (int i = 0; i < state_string.length(); i++)
    {

        if ((ignored_pancakes.find(state_string[i + 1]) != string::npos) or (ignored_pancakes.find(state_string.at(i)) != string::npos))
        {
            continue;
        }

        if (abs(goal_state_string.find(state_string[i])-goal_state_string.find(state_string[i+1])!=1)){
            gap++;
        }


        cout << state_string.at(i) << "\t" << state_string.at(i + 1) << endl;
    }
    // cout << state_string << endl;
    cout << ignored_pancakes << endl;
    cout << gap << endl;
 }

我期望的输出如下：

3       4
4       5
5       6
012
2

但是打印出来的是：

3       4
4       5
5       6

奇怪的是，当我注释掉那行：

cout << state_string.at(i) << "\t" << state_string.at(i + 1) << endl;

输出为：

012
2

为什么根据那行打印出完全不同的内容，这似乎无关紧要。

最佳答案

for (int i = 0; i < state_string.length(); i++)
{
    // ...
    cout << state_string.at(i) << "\t" << state_string.at(i + 1) << endl;
}

使用i + 1您将访问字符串超出范围。将循环更改为

for (int i = 0; i < state_string.length() - 1; i++)

关于c++ - cpp此功能的异常行为，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/52529210/