我一直在研究的这个c ++程序正在生成涉及以下内容的运行时错误：
1.generatePass方法跳过验证过程。
2.validatePass方法会跳过用户输入并跳过验证。
我是c ++的新手，是否有某种类型的nextLine方法可以添加到java中的用户输入以及为什么要跳过验证的任何想法。不同的视角会有所帮助，因为我还没有找到任何其他内容远。
感谢布。
#include <stdlib.h>
#include <stdio.h>
#include <iostream>
#include <string>
#include <windows.h>
#include <ctime>
#include <string.h>
#include <time.h>
using namespace std;
#define MAX 80
//declaring functions
int showMenu();
void generatePass();
void validatePass();
int countLetters(char *,int *,int *,int *,int *,int *);
int main()
{
int iChoice;
// have menu appear, user makes decision, do work, reshow menu
// do this until user enters 5
do
{
iChoice = showMenu();
}while(iChoice != 3);
printf("\n\n\n");
system("pause");
}//end of main
//Methods placed here:
//showMenu method calls program menu,either 1.generate password,2.enter password and validate. or 3.exit(close program)
int showMenu()
{
int iChoice;
system("cls");
printf("\n\n\t\tWelcome to Password Generator and Validator\n\n");
printf("\n\t\t1. Generate");
printf("\n\t\t2. Validate");
printf("\n\t\t3. Exit");
printf("\n\n\t\tEnter your menu choice: ");
fflush(stdin);
scanf_s("%d", &iChoice);
// user enters one of 3 values
// generate,validate or exit program
switch(iChoice)
{
case 1: // generate
{
generatePass();
break;
}
case 2: // validate
{
validatePass();
break;
}
case 3: // exit
{
printf("\n\nProgram exiting!...");
break;
}
default:
{
break;
}
}//end of switch
return(iChoice);
} //end of showMenu
//method to generate a random password for user following password guidelines.
void generatePass()
{
int iChar,iUpper,iLower,iSymbol,iNumber,iTotal;
printf("\n\n\t\tGenerate Password selected ");
printf("\n\n\t\tPassword creation in progress... ");
int i;
char password[10 + 1];
char strLower[59+1] = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRTUVWXYZ!£$%^&*";
srand(time (0));
for(i = 0; i < 10;i++)
{
password[i] = strLower[(rand() % 52)];
}
password[i] = '\0';
iChar = countLetters(password,&iUpper,&iLower,&iSymbol,&iNumber,&iTotal);
if(iUpper < 2)
{
printf("Not enough uppercase letters!!!\n");
}
else if(iLower < 2)
{
printf("Not enough lowercase letters!!!\n");
}
else if(iSymbol < 1)
{
printf("Not enough symbols!!!\n");
}
else if(iNumber < 2)
{
printf("Not enough numbers!!!\n");
}
else if(iTotal < 9 && iTotal > 15)
{
printf("Not enough characters!!!\n");
}
printf("\n\n\n Your new password is verified ");
printf(password);
printf("\n\n\n");
system("pause");
}//end of generatePass method.
//method to validate a user generated password following password guidelines.
void validatePass()
{
char password[MAX+1];
int iChar,iUpper,iLower,iSymbol,iNumber,iTotal;
//shows user password guidelines
printf("\n\n\t\tPassword rules: ");
printf("\n\n\t\t 1. Passwords must be at least 9 characters long and less than 15 characters. ");
printf("\n\n\t\t 2. Passwords must have at least 2 numbers in them.");
printf("\n\n\t\t 3. Passwords must have at least 2 uppercase letters and 2 lowercase letters in them.");
printf("\n\n\t\t 4. Passwords must have at least 1 symbol in them (eg ?, $, £, %).");
printf("\n\n\t\t 5. Passwords may not have small, common words in them eg hat, pow or ate.");
//gets user password input
printf("\n\n\t\tEnter your password following password rules: ");
gets_s(password);
iChar = countLetters(password,&iUpper,&iLower,&iSymbol,&iNumber,&iTotal);
if(iUpper < 2)
{
printf("Not enough uppercase letters!!!\n");
}
else if(iLower < 2)
{
printf("Not enough lowercase letters!!!\n");
}
else if(iSymbol < 1)
{
printf("Not enough symbols!!!\n");
}
else if(iNumber < 2)
{
printf("Not enough numbers!!!\n");
}
else if(iTotal < 9 && iTotal > 15)
{
printf("Not enough characters!!!\n");
}
printf("\n\n\n Your new password is verified ");
printf(password);
printf("\n\n\n");
system("pause");
}//end validatePass method
int countLetters(char * Password,int * Upper,int * Lower,int * Symbol,int * Number,int * Total)
{
int iTotal = 0,iC = 0,tU = 0,tL = 0,tS = 0,tN = 0;
//strlen- function that returns length
for (int iC = 0;iC < strlen(Password);iC++)
{
printf("%d",Password[iC]);
//uppercase letters are in the range 65 - 90
//lowercase letters are in the range 97 - 122
//symbols are in the range 32-48
//numbers are in the range 47 - 58
if((Password[iC] < 64) && (Password[iC] < 91))
{
tU++;
iTotal++;
}
else if((Password[iC] > 96) && (Password[iC] < 123))
{
tL++;
iTotal++;
}
else if((Password[iC] > 32) && (Password[iC] < 48))
{
tS++;
iTotal++;
}
else if((Password[iC] > 47) && (Password[iC] < 58))
{
tN++;
iTotal++;
}
*Upper = tU;/*set value at memory address = tU,passing by reference saves memory used.*/
*Lower = tL;
*Symbol = tS;
*Number = tN;
}//end for statement
return (iTotal);
}//end of countLetters
答案 0 :(得分：1)
我认为主要的问题是，虽然你有构建模块，但你还没有完全跟上规划你的设计。别担心，这是可以解决的。
1.generatePass方法跳过验证过程。
generatePass（）在其自身内执行验证，但您可能想要放置这些行......
printf("\n\n\n Your new password is verified ");
printf(password);
...进入一个else块，这样用户就不会认为密码没问题。你也可能想利用generatePass（）中的validatePass（）函数，因为你正在重复代码。
2.validatePass方法会跳过用户输入并跳过验证。
也许此链接与您使用gets_s（）：StackOverflow.com时遇到的问题相关。无论如何，你可以暂时使用scanf，因为它一开始会让事情变得更容易，如果你需要，你可以稍后返回并使程序更加强大。 Scanf将修复被绕过的输入。跳过验证与＃1的模式相同，所以如果你弄清楚你想要做什么，那么这里很容易。
我不确定我是否理解Java nextLine（）问题，但使用带有％s的scanf来读取与Java Scanner.nextLine（）大致类似的字符串函数。
免责声明：我将您的程序作为C程序运行，因为将它从C ++转换为C是微不足道的。（我想如果您要编写C风格的代码，您可能会考虑将其设为C语言类似地，你可以选择避免仅使用Windows的东西（例如，'_s'函数），如果你想让程序更便携，但你可能会从'_s'函数中获得一些好的实用程序。重做，所以这只是一个想法。）