Creating a simple crackme program in C++ problems with variable and input - c++

Question

So I am trying to learn C++ so I can learn some reverse engineering that why I am trying to create this simple crack me program to get a foundation going and not take someone else's project as I choose my own path. However I am using CodeBlocks as the other IDEs were not being cooperative and am enjoying it and has given me some error and two lines. Below is the following code. So there errors are the following:
||=== Build: Debug in SimpleProgram (compiler: GNU GCC Compiler) ===|
D:\Programming\C++ Projects\simple programs\SimpleProgram\main.cpp||In member function 'int checker::processing(int)':|
D:\Programming\C++ Projects\simple programs\SimpleProgram\main.cpp|15|warning: no return statement in function returning non-void [-Wreturn-type]|
D:\Programming\C++ Projects\simple programs\SimpleProgram\main.cpp||In function 'int main()':|
D:\Programming\C++ Projects\simple programs\SimpleProgram\main.cpp|22|error: 'x' was not declared in this scope|
||=== Build failed: 1 error(s), 1 warning(s) (0 minute(s), 0 second(s)) ===|
#include <iostream>
using namespace std;
class checker{
public:
int number;
processing(int x){
x = number;
if ( number == 10 ){
cout << "Well done!";
} else {
cout << "Keep trying!";
}
}
};
int main()
{
checker cracking;
cout << "Please enter in the correct number";
cin >> cracking.processing(x);
return 0;
}
Image of the project and error

Answer 1

A function always has a return type, even if your not attempting to return anything it will have the void signature. If your intent was to take input of a number passed from main and display it via a function in your class checker through an object of the same, this is how it would look like:
#include <iostream>
using namespace std;
class checker{
public:
int number;
void processing(int x)
{
if (x==10)
cout << "Well done!";
else
cout << "Keep trying!";
}
};
int main()
{
checker cracking;
cout << "Please enter in the correct number \n";
int n;
cin >> n;
cracking.processing(n);
return 0;
}

Answer 2

I've cleaned the code up and included comments that serve as notes:
#include <iostream>
using namespace std;
class checker{
public:
void setnumber(int i){ //it's a good habit to put variables in private and access them with a public function
this->number = i;
};
int processing(int x){ //x is just a placeholder value for whatever you put in here. You can't use it in the rest of the program
if ( x == 10 ){
cout << "Well done!" << endl;
return 1; //this condition indicates success
} else {
cout << "Keep trying!" << endl; //the endline just makes it so you aren't typing on the same line as the message
return 0; //this condition indicates success hasn't been reached yet
}
}
private:
int number;
};
int main()
{
checker cracking;
cracking.setnumber(10); //the number is set to 10
int i, l; //i is the guess number, l is a condition for the loop
cout << "Please enter in the correct number" << endl;
do{ //this loop sets it up so that you can have multiple tries
cin >> i;
l = cracking.processing(i);
}while(l!=1); //and this condition (the return value of processing(), breaks the loop on success
return 0;
}
The main issue that popped out at me was the use of x.
Trying to set x to number. In functions, the parameters are just placeholder values for arguments that will be passed into later. Then later on when you tried to use x as an input in the main() program. You were calling that function (using it) and needed an int as input.
Don't worry. It's confusing in the beginning for everyone (although to be fair, as you progress you'll just find new things to be confused about. It never really stops). Keep at it and it'll all make sense in time.