用变量和输入在C ++问题中创建一个简单的Crackme程序

Question

所以我试图学习C ++，所以我可以学习一些逆向工程，这就是为什么我试图创建这个简单的crack me程序来获得基础，而不是在选择自己的道路时接受别人的项目。但是，我正在使用CodeBlocks，因为其他IDE并不协作，我很享受它，并且给了我一些错误和两行代码。以下是以下代码。因此，错误如下：

||=== 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

函数始终具有返回类型，即使您不尝试返回任何内容，该函数也会具有void签名。如果您打算输入从main传来的数字，并通过类检查器中的函数通过相同的对象显示该数字，则它看起来像这样：

#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

我已经清理了代码，并包含了用作注释的注释：

#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;
}

向我冒出的主要问题是使用x。

尝试将x设置为number。在函数中，参数只是参数的占位符值，这些参数将在以后传递给参数。然后，当您尝试在main()程序中使用x作为输入时。您正在调用该函数（使用它），并且需要一个int作为输入。

不用担心。在一开始，每个人都会感到困惑（虽然公平地说，随着您的进步，您会发现新事物容易混淆。它永远不会停止）。坚持下去，一切都会及时生效。