有人可以帮助我实现这一目标吗？ Thx提前。 [英] Can someone help me make this come true ? Thx in advance .

问题描述

所以最近我决定制作一个密码生成器，提示大量的char。在密码中，然后提示我遇到的任何类型的字符（az，AZ，0-9，还有一些和自定义的字符），然后输出所有可能的
密码组合到txt文件。然而我无法找到解决方案，请帮助我，提示和调试意味着很多。至于我已经得到的错误，他们都是不同的没有错误，告诉我，我有类型不匹配，
但它在某种形式的代码，如堆栈溢出一，这花了我一天了解。我会上传我的流程，所以你可以亲自看看，proj 1

So recently i decided to make a password generator that , prompts for amount of char. in a password , then prompts for whatever type of char(a-z , A-Z , 0-9 , and some more and a custom one ) that i had a hard time with , which then outputs all the possible combinations of the password to a txt file . And yet i cant find the solution , please help me , tips and debugging would mean much. As for the errors i've been getting , well they are all different no errors that tell me that i have a mismatch in types , but its in some form of code like the stack overflow one , that took me a day to understand . I will upload my process ,so you can see for yourselves, proj 1

#include "stdafx.h"
#include <iostream>
#include <fstream>
#include <cmath>
#include <string>
#include <stdlib.h>
#include <istream>
using namespace std;
int main()
{
	ofstream THEOUTPUT("BLYAT.txt");
	string S1[26]; string S2[26]; string S3[26]; string S4[26]; string S5[26]; string S6[26]; string S7[26]; string S8[26]; string S9[26]; string S10[26]; string S11[26]; string S12[26];
	string PASSWORD[6000]; string SHITPASS[6000]; int n = 0; int m = 0; int x = 0; int y = 0; int counter = 1; string SS;
	char chars[26];//small , caps , numbers
	int z = 0; string SSS; string SSSS; int k = 0;
	char specificchars[100];
	int lol = 0;
	while (1<5)

	{
		for (int a = 9; a<123; a++)
		{
			int counter1 = a - 97;
			S1[counter1] = (char)a;
			cout << S1[counter1] << endl;
		}
		for (int a = 97; a<123; a++)
		{

			int counter2 = a - 97;
			S2[counter2] = (char)a;
			cout << S2[counter2] << endl;
		}
		for (int a = 97; a<123; a++)
		{
			int counter3 = a - 97;
			S3[counter3] = (char)a;
			cout << S3[counter3] << endl;
		}
		while (counter != 2000)
		{
	
			if (S2[lol] == "y" && S3[z] == "z") lol++;
			PASSWORD[n] = S1[x] + S2[lol] + S3[z];

			if (S2[lol] == "z" && S3[z] == "z")
			{
				x++;
			}
			if (S2[lol] == "y" && S3[z] == "z")
			{
				z = 0;

			}
			if (S3[z] == "z")
			{
				lol++;
				k++;
				
			}
			if (k == 1)
			{
				z = -1;
				k = 0;
			}

			if (S2[lol] == "z")
			{
				x++;
				lol = 0;
			}
			z++;


			THEOUTPUT << counter << " " << PASSWORD[n] << endl;
			n++;
			counter++;
			if (S1[x] == "z" && S2[lol] == "z" && S3[z] == "z")//exit
			{
				break;
			}

		}

		//	PASSWORD[n]=S1+S2+S3+S4+S5+S6+S7+S8+S9+S10+S11+S12;	
	}
	system("PAUSE");
	return 0;
	THEOUTPUT.close();
}

因此，我所有努力的要点归结为这段代码，但是我试图让它发生它永远不会发生，那就产生了密码来自26个小字符，每个单独的3个字符位置。再次提前致谢。

So the gist of all my efforts comes down to this piece of code , however i try to make it happen it never does, thats generates the passwords from 26 small characters , on 3 character places for every separate one. Thanks in advance again.

推荐答案

	while (1<5)

	{
		for (int a = 9; a<123; a++)
		{
			int counter1 = a - 97;
			S1[counter1] = (char)a;
			cout << S1[counter1] << endl;
		}
		for (int a = 97; a<123; a++)
		{

			int counter2 = a - 97;
			S2[counter2] = (char)a;
			cout << S2[counter2] << endl;
		}
		for (int a = 97; a<123; a++)
		{
			int counter3 = a - 97;
			S3[counter3] = (char)a;
			cout << S3[counter3] << endl;
		}
		while (counter != 2000)
		{
	
			if (S2[lol] == "y" && S3[z] == "z") lol++;
			PASSWORD[n] = S1[x] + S2[lol] + S3[z];

			if (S2[lol] == "z" && S3[z] == "z")
			{
				x++;
			}
			if (S2[lol] == "y" && S3[z] == "z")
			{
				z = 0;

			}
			if (S3[z] == "z")
			{
				lol++;
				k++;
				
			}
			if (k == 1)
			{
				z = -1;
				k = 0;
			}

			if (S2[lol] == "z")
			{
				x++;
				lol = 0;
			}
			z++;


			THEOUTPUT << counter << " " << PASSWORD[n] << endl;
			n++;
			counter++;
			if (S1[x] == "z" && S2[lol] == "z" && S3[z] == "z")//exit
			{
				break;
			}

		}

		//	PASSWORD[n]=S1+S2+S3+S4+S5+S6+S7+S8+S9+S10+S11+S12;	
	}
	system("PAUSE");
	return 0;
	THEOUTPUT.close();
}

对于初学者我建议你设置警告等级4并进行重建。

然后纠正任何问题。



> while（1< 5）



因为这将永远是真的，你如何摆脱这个循环？



我看到一个打破;声明，但这是在另一个while（）循环内，所以在
它将导致内循环退出。但是这个外循环呢？
$


>返回0;

> THEOUTPUT。 close（）;



返回语句之后的任何内容都不会被执行，因为它永远无法到达。



- Wayne

For starters I suggest that you set Warning Level 4 and do a rebuild.
Then correct any issues.

>while (1<5)

Since this will always be true, how do you get out of this loop?

I see one break; statement, but that is inside another while() loop so
it will cause that inner loop to exit. But what about this outer loop?

>return 0;
>THEOUTPUT.close();

Nothing after a return statement will ever be executed as it can never be reached.

- Wayne

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