随机密码生成器的C++循环错误"以非零状态退出"

C++ Loop Error "Exited With Non-Zero Status" for Random Password Generator

本文关键字：退出 状态 错误 循环 密码 C++ 随机      更新时间：2023-10-16

因此，我正在尝试在C 中学习循环和条件，因此我决定编写一个为用户生成随机密码的程序。由于某种原因，该代码可能工作1/5次，但其余时间才使我"以非零状态退出"。谢谢，evin 

#include <iostream>
#include <cstdlib>
#include <chrono>
#include <thread>
#include <time.h>
using namespace std;
int main() 
{
    using namespace std::this_thread;
    using namespace std::chrono;
//  Vars
    string lett;
    int input;
    string password("");
    string lettArray [] = {"a", "b", "c", "d", "e","f", "g", "h", "i", "j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"};
//  Prompt
    cout << "How long would you like your password to be?";
    cin >> input;
//  Loop
    for(int i = 0; i < input; i++)
    {
        struct timespec ts;
        clock_gettime(CLOCK_MONOTONIC, &ts);
        srand((time_t)ts.tv_nsec);
        int random = (rand() % 26 + 1);
        lett = lettArray[random];
        password = password + lett;
        sleep_for(milliseconds(10));
        cout << "." << endl;
        if (random == 0 )
            break;
    }
//  Output
    cout << password << endl;
    return 0;
}

此程序随机失败（双关语）是因为您使用的是c-arrays，并且索引范围错误。

c-array，例如lettarray，请勿检查是否违反了数组界限。在您的示例中，如果运行此程序，lettArray[26]将为您提供分割故障，因为该内存地址中没有string元素。C阵列没有检查界限，因此很难知道出了什么问题。这在复杂的程序中可能会变得特别棘手，因为如果碰巧在该内存地址中有某些东西，您可以获得毫无意义的结果。

更好的实现是使用std::vector：

#include <iostream>
#include <cstdlib>
#include <chrono>
#include <thread>
#include <time.h>
#include <vector>
using namespace std;
int main() 
{
    // You don't need these two lines
    //using namespace std::this_thread;
    //using namespace std::chrono;
    //  Vars
    string lett;
    int input;
    string password("");
    // Vector index range: 0-25
    vector<string> lettArray = {"a", "b", "c", "d", "e","f", "g", "h", "i", "j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"};
    // You only need to initialise the seed once.
    struct timespec ts;
    clock_gettime(CLOCK_MONOTONIC, &ts);
    srand((time_t)ts.tv_nsec);
//  Prompt
    cout << "How long would you like your password to be?";
    cin >> input;
//  Loop
    for(int i = 0; i < input; i++)
    {
        int random = (rand() % 26 ); // You get results between 0-25
        // Using lettArray.at(26) will result in a crash with a clear 
        // message that bounds are violated  
        lett = lettArray.at(random); 
        password = password + lett;
        cout << "." << endl;
       // Don't see a reason for this if statement
       // if (random == 0 ){
       //     continue;
       //}
    }
//  Output
    cout << password << endl;
    return 0;
}

我还将随机种子移到循环外（您只需要一次），也没有理由将10毫秒的暂停。您会看到您的结果是正确随机的。

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