C ++循环错误"退出非零状态"用于随机密码生成器

时间:2016-12-09 18:26:08

标签: c++ arrays loops compiler-errors conditional

所以我试图在c ++中学习循环和条件,所以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,代码可能工作了1/5倍,但其余的时间只是给了我"退出非零状态"。 谢谢, 的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;
}

1 个答案:

答案 0 :(得分:0)

这个程序随机失败的原因(双关语意思是)你正在使用C数组,并且索引范围错误。

C数组(例如lettArray)不检查数组边界是否被违反。在您的示例中,如果您运行此程序,lettArray[26]将为您提供分段错误,因为该内存地址中没有string元素。 C阵列不检查边界,因此很难知道出了什么问题。在复杂的程序中,这可能会变得特别棘手,因为如果某个内存地址恰好出现在内存地址中,您可能会得到无意义的结果。

更好的实施方法是使用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了循环之外(你只需要这一次),也没有理由暂停10毫秒。无论如何,您都会看到您的结果是正确随机的。