I have been trying to make a brute-force program in C++, for a school project, but I want to make the password you enter censored, so instead of like password it would be **
PS it's an console application
This is my C++ Code:
#include <iostream>
#include <string>
using namespace std;
char chars[]={'0','1','2','3','4','5','6','7','8','9','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','A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','W','R','S','T','U','V','W','X','Y','Z'};
string t;
void checkPassword(string password);
void recurse(int width, int position, string baseString);
int main() {
cout << "Enter a Password: " << endl;
cin >> t;
int maxChars = 13;
for(int i=0;i<maxChars+1;i++) {
cout << "checking passwords width [" << i << "]..." << endl;
recurse(i,0,"");
}
return 0;
}
void recurse(int width, int position, string baseString) {
for(int i=0;i<70;i++) {
if (position < width-1) {
recurse(width, position + 1, baseString+chars[i]);
}
checkPassword(baseString+chars[i]);
}
}
void checkPassword(string password) {
if (password==t) {
cout << "Match Found: " << password << "" << endl;
return;
}
}
I have tried a lot but I can't seem to get it working, thanks in advance.
A console program doesn't have the capabilities to do this. If you're working on a Windows machine and don't need any cross-platform portability, you may be able to get away with using the non-standard conio.h library's getch() function to simulate cin while replacing the characters. But I honestly recommend against a roundabout approach like that. If you really need the censorship of the password, I suggest that you either turn the program into a GUI application or ask the user to type their password into a file and then give the filename (so that it doesn't appear onscreen as directly).
Related
I am new in VS code. I wrote a C++ code like one below. but unfortunately in the terminal or output panel I cannot get both of the string and variable value. in the terminal only variable's inputted value is showing. How to fix this?
#include <bits/stdc++.h>
int main()
{
int slices;
std::cin >> slices;
std::cout << "You got " << slices << " of pizzas" << std::endl;
return 0;
}
Hello guys i am beginner on the language c++
i was trying to run this code below on my ide"codeblocks" and it works
https://www.youtube.com/watch?v=vLnPwxZdW4Y (link for the tutorial that following )
#include <iostream>
using namespace std;
int main()
{
string charactername = "arnold";
int characterage;
characterage = 10;
cout << "Hello my name is" << charactername<< endl;
cout << "i am " << characterage << endl;
return 0;
}
this code does not work on my other compiler running on dosbox ? any ideas why ?
I suggest you stop using Turbo C++ as it is a very outdated and a discontinued compiler. However, if you don't have the option of using new compilers (I had the same issue as I studied C++ at school), you will have to make the following changes:
using namespace std; cannot be used in Turbo C++. You will have to remove that and replace #include<iostream> with #include<iostream.h>
Data-type string cannot be used in Turbo C++. You will have to declare a character array instead.
You will have to use #include<stdio.h> and the function puts(); to display the character array in case of Turbo C++. Alternatively you can use a loop-statement.
This will be your final code:
#include <iostream.h>
#include <stdio.h>
int main()
{
char charactername[] = "arnold";
int characterage;
characterage = 10;
cout << "Hello my name is ";
puts(charactername);
cout << "i am " << characterage << endl;
return 0;
}
Note: The puts(); function automatically puts the cursor on the next line. So you don't need to use endl;
Or, if you want to use a loop-statement to display the character array
#include <iostream.h>
int main()
{
char charactername[] = "arnold";
int characterage;
characterage = 10;
cout << "Hello my name is ";
int i=0;
while(charactername[i]!='\0') {
cout<<charactername[i];
i++;
}
cout<<endl;
cout << "i am " << characterage << endl;
return 0;
}
'\0' is the last element of the character array. So as long as the loop does not reach the last element, it will print the character array.
a[] = "arnold"; basically means an array is created like this: a[0]='a', a[1]='r', a[2]='n',.... a[5]='d', a[6]='\0'.
Alternatively, if you use cout << charactername; instead of the while loop, it will print the whole name. (This is only in the case of a string variable (character array), for an integer array or any other array you will need the while loop)
I am receiving the following debug error when attempting to run the first part of my program:
Debug Error!
Program:
...\user\desktop\PunchLineProgram\Debug\PunchLineProgram.exe
Module:
...\user\desktop\PunchLineProgram\Debug\PunchLineProgram.exe
File:
Run-Time Check Failure #3 - T
(Press Retry to debug the application)
I am attempting to have the user select whether they want to hear a joke, running and if\else statement that will output a message to the user, based on their response. If I comment out these statements, I do not receive the error when attempting to run the program. I know I'm probably missing something simple, as I am a novice. Here is the code that I have so far:
/*Include Section*/
#include <iostream>
#include <string>
#include <fstream>
#include <iomanip>
#include <cctype>
/*Namespace Section*/
using namespace std;
/*Function Prototypes Section*/
void displayAllLines(ifstream &infile);
void displayLastLine(ifstream &infile);
/*Main section: this is the entry point of the program, which controls the flow of execution*/
int main()
{
string file1;
string file2;
ifstream joke;
ifstream punchline;
int decision;
char y;
char n;
cout << "*******************************************************************************" << endl;
cout << setw(48) << "Punchline Program" << endl;
cout << "*******************************************************************************" << endl;
cout << endl;
cout << "Welcome to the Punchline Program!" << endl;
cout << "Are you ready to hear a joke? (y or n): ";
cin >> decision;
if (decision == y)
{
cout << "Great! Let's get started!" << endl;
}
else if (decision == n)
{
cout << "Ah, no sense of humor, I see. Time to make like a tree and leaf (queue rimshot)!" << endl;
}
system("PAUSE");
}
Any help would be greatly appreciated!
When comparing to char you should use '':
char answer
if (answer == 'y') { *//this only checks for LOWER case y*
cout << "You selected Yes" << endl;
}
when comparing to a string use ""
int/float/double... you can just use the variable.
Besides that, your decision variable as int when it should be char, and you don't need char y nor n. (you yourself never even used it in that code)
I'd suggest looking up c++ tutorials, most show and explain the different between char/string, ' and ".
I am fairly new to C++ and coding in general. I am trying to make just a basic little multiple choice type game for practice but I have run into a conundrum.
The program isn't outputting what I want it too. Here is the code:
#include <iostream>
#include <cstdlib>
#include <ctime>
#include <string>
using namespace std;
void sword(int damage);
void fists(int damage);
static int enemyHealth = 250;
int main() {
srand(time(0));
string SOF; //Abreveation for "Sword Or Fists"
cout << "You will be fighting a mean bad guy. Are you using a sword, or your fists?\n";
while (SOF != "sword" && SOF != "fists"){
cout << "Please enter your choice of either 'sword' or 'fists': ";
cin >> SOF;
}
cout << "Okay! Time to fight! \n";
if (SOF == "fists") {
void fists();
}
else if (SOF == "sword") {
void sword();
}
else{ (NULL); }
cout << "Congratulations! You have vanquished that foul beast!\n";
system("pause");
}
//This is for when the user chooses 'sword'
void sword(int damage = rand() % 100 + 50) {
while (enemyHealth > 0){
cout << "You deal " << damage << " damage with your sharp sword. \n";
enemyHealth -= damage;
}
}
//This is for when the user chooses 'fists'
void fists(int damage = rand() % 10 + 4) {
while (enemyHealth > 0){
cout << "You deal " << damage << " damage with your womanly fists. \n";
enemyHealth -= damage;
}
}
The first part works fine, but when I enter my choice of either "fists" or "sword" the output is:
Okay! Time to fight!
Congratulations! You have vanquished that foul beast!
But I want it to output the damage being done with either fists or sword.
If I could get some help with that, it would be amazing. Thanks!
void fists(); is a declaration, not a call, change to fists(); and sword();
Other things to look at:
Default parameters are declared in function declaration before main (or just move whole functions there)
Default parameters in c++ are evaluated once, so all 'hits' will be the same in your code
Local variable names are usually not named in uppercase, SOF looks loke it is a #defined constant or such.
To call the function, don't write void fists();, just
fists();
(What you have is a declaration, which has no useful effect here, rather than a call.)
I have tried to adapt my knowledge of modularity to Visual C++ however, upon what seems to be an endless search scouring for syntax, I simply can't get this right. Basically in this code, the menu is called first, once the user enters their choice (only coded option 1 thus far) to return that value to the main, which then steps into the if statement and calls fahrenheit. I am requesting the syntax for passing by reference, I know C#'s syntax for this, but not Visual C++
Here's the code.
// Test.cpp : Defines the entry point for the console application.
//
#include "stdafx.h"
#include <iostream>
#include <string>
using namespace std;
void Celsius()
{
}
void fahrenheit()
{
cout << "Success!" << endl; //....Outputs this just to see if the module is being called properly.
}
int menu(int Mystring) //....I was testing this syntax to pass the variable.
{
cout << "What would you like to do : " << endl;
cout << "1) Fanreheit to Celsius" << endl;
cout << "2) Celsius to Fahrenheit" << endl;
cout << "Choice : " ;
cin >> Mystring;
return Mystring;
}
int main()
{
int celsius = 0;
int fahrenheit = 0;
int Mystring = 0;
menu(Mystring); //....Testing this syntax to pass Mystring to menu.
if (Mystring == 1) //....I was hoping the menu would return Mystring as value = 1.
{
fahrenheit(); //.......I want this to call fahrenheit module if Mystring = 1
}
}
The "things" you're talking about aren't called modules, but functions. That's a pretty big difference and I think you should know it, since you won't understand nearly any article without that knowledge.
That being cleared, the problem in your code is, that you pass the variable by value (int menu(int Mystring)), while - in order to change it inside the function - you need to pass it by reference or pointer:
int menu(int &Mystring)
There are plenty of articles about functions in C++. You should check them out probably.