out_of_range在内存位置c ++

Question

在该程序中,用户在命令行中输入两个字符串.其中一个是数字,另一个是字符串.该程序用于根据电话拨号盘上每个按钮上的数字/字母检查输入的数字和字符串是否相同.我得到一个例外,说:

Unhandled exception at 0x775DDAE8 in Project10.exe: 
Microsoft C++ exception: std::out_of_range at memory location 0x0018F158. 

非常感谢任何帮助,非常感谢你:)

using namespace std;

bool checkPswd(string keyStrokes, string password) {
    string temp;
    string temp2;
    bool temporary = 1;
    string phoneButtons[10] = {
        "", ""
        "abc", "def", "ghi", "jkl",
        "mno", "pqrs", "tuv", "wxyz"
    };
    for (int i = 0; i < keyStrokes.length(); i++) {
        for (int j = 2; j < 10; j++) {
            for (int k = 0; k < phoneButtons[j].length(); k++) {
                temp = phoneButtons[j];
                if (password.at(k) == temp.at(k)) {
                    temp2 = +(char)k;
                }
                else {
                    //do nothing
                }
            }
        }
    }
    cout << temp2;
    for (int m = 0; m < temp2.length(); m++) {
        if (temp2.at(m) == keyStrokes.at(m)) {
            //keep searching
        }
        else {
            return 0;
        }
    }
    return 1;
}

int main(int argc, char ** argv) {
    if (argc != 3) {
        cout << "Please input the key strokes from the phone and the password." << endl;
        return 1;
    }

    string keyStrokes = argv[1];
    string password = argv[2];


    bool check;

    check = checkPswd(keyStrokes, password);
    if (check) {
        cout << "Password Verified" << endl;
    }
    else {
        cout << "Wrong Password" << endl;
    }
    return 0;
}

Answer 1

for (int k = 0; k < phoneButtons[j].length(); k++) {

k将在这里迭代,从0到1小于`phoneButtons [j]中的字符数

   if (password.at(k)  // ... rest of the code is irrelevant

流行测验:如果password字符少于phoneButtons[j],你认为这会发生什么？

for (int m = 0; m < temp2.length(); m++) {
    if (temp2.at(m) == keyStrokes.at(m)) {

同样的错误也在这里.我还没有分析所显示代码的其余部分,以了解同一逻辑错误的任何其他情况.

这就是你应该使用的原因#include <algorithm>,而不是使用这些容易出错的C风格for循环.std::find()比这种基于C风格的循环搜索更不容易出错.