C ++中的意外无限循环

Question

我正在一个项目中，任务是进行注册功能因此，当我在密码功能中尝试多个测试用例时，当我尝试输入无效的密码而没有给我重新输入密码的机会时，我陷入了无限循环。预先感谢...

#include <iostream>
#include <string>
#include <cstring>
#include <conio.h>
#include <stdio.h>
using namespace std;
#define size 40
int main(){
    bool flag = true;
    while (flag){
        bool flagS = false, flagN = false; int count = 0;
        char password[size] = { 0 };
        cout << "Type your Password .....\n Note :: Must be more than 8 characters including at least one number and one special character...\n";
        cin.get(password, size);
        count = strlen(password);
        for (int z = 0; z < count; z++){
            if (password[z] >= 48 && password[z] <= 57)
                flagN = 1;
            if ((password[z] >= 33 && password[z] <= 47) || (password[z] >= 58 && password[z] <= 64))
                flagS = 1;
        }

        if ((flagS == 1) && (flagN == 1) && (count >= 8))
        {
            cout << "Valide Password ...\nCongrats!! ..you created a NEW account.." << endl;
            flag = false;
        }
        else
        {
            cout << "invalide password..\nPlease try again..\n";
            flag = true;
        }
    }
}

Answer 1

您必须在cin循环内写入while。这将解决您的问题：

#include <iostream>
#include <cstring>
#include <stdio.h>
using namespace std;
#define size 40
int main(){
    bool flag = true;
    bool flagS = false, flagN = false; int count = 0;
    char password[size] = { 0 };

    cout << "Type your Password .....\n Note :: Must be more than 8 characters including at least one number and one special character...\n";

    while (cin >> password && flag){

        count = strlen(password);
        for (int z = 0; z < count; z++){
            if (password[z] >= 48 && password[z] <= 57)
                flagN = true;
            if ((password[z] >= 33 && password[z] <= 47) || (password[z] >= 58 && password[z] <= 64))
                flagS = true;
        }

        if ((flagS) && (flagN) && (count >= 8))
        {
            cout << "Valide Password ...\nCongrats!! ..you created a NEW account.." << endl;
            flag = false;
        }
        else
        {
            cout << "invalide password..\nPlease try again..\n" << "Type your Password .....\n Note :: Must be more than 8 characters including at least one number and one special character...\n";;
            flag = true;
        }
    }
}

Answer 2

std::istream::get不会从流中删除换行符。这意味着对std::istream::get的下一次调用会立即找到换行符，而不将任何内容读入缓冲区，设置故障位，然后返回。因为现在立体声处于失败状态，所以所有后续读取都会立即失败。您可以std::istream::get换行符，但是更直接的方法是使用cin.ignore()，因为它会为您删除换行符。

始终在IO事务后测试流状态以确保成功。

旁注：

[首选将std::istream::getline和std::istream::getline用于原始字符数组和使用它们的函数。如果提供的缓冲区太小，std::string和std::getline都将流标记为失败。提供给std::istream::get的std::istream::getline将被调整大小，因此，只要您有剩余的动态存储，就可以解决缓冲区溢出问题以及为防止这些问题而引发的错误。如果系统无法为string分配足够大小的缓冲区，则会引发异常以使您意识到问题。

[首选使用字母，例如std::getline代替原始ASCII码。一方面，即使最原始的程序员也可以立即认识到string的意图，并且始终有可能实现不使用ASCII作为默认编码。