Decode a message without a key

Question

Here is what I should do making a program that reads a text file that contains encrypted message and crack it it is kind of close to substitution cipher where I should swap swap each letter back with another meaning like shifting B back to A if its being shifted by one, and try comparing shifted words by some common used words to find if 2 of the common words have been found on the shifted ones

ex: shift the word by one
check when you shifted it is there 2 common words found?
no keep going
yes means thats it stop shifting

Now this issue here which makes the program hard for me is that I do not have a key to be entered it would've been awesome if I had it.

3 issues I am having now, is in my void function it will not print anything, the second issue is, even if I fix my issue (I know this because in my function I have added something to convert string to c_string) and the issue is it will not shift until it finds the words I am looking which I have declared in string "Common[]" 3rd issue is whenever I compile I get an error invalid conversion from const char** to char comparison between signed and unsigned integer expressions

#include <iostream>
#include <fstream>
#include <string>
#include <stdio.h>
#include <ctype.h>
using namespace std;

void makeshift (string encrypt, int key)
{
    for (int i = 0 ; i<encrypt.size();i++)
    {
        cout<<encrypt<<endl; // to see what is in file way to debug
        const char *encryptedc; // this is to convert string to cstring
        encryptedc = encrypt.c_str();
        encryptedc-=key;
        cout<<"decrypted message is "<<encryptedc<<endl;
    }
}


int main (int argc, char* argv[])
{
    // this will make me compare between encrypted message to check if it
    // contains this words or not!
    const char* common[]{"the","and","in","he","with","that","as","at","do"};
    
    string encrypted; // The encrypted message gotten from file
    
    ifstream message(argv[1]); // get file name typed by user
    if(message.is_open()) // check if file is open
    {
        while (!message.eof()) // check if we reached end of file
        {
            // get the whole line in file
            getline(message,encrypted);
            
            ///loop throgh it to store numbers in declared varibles
            for (int i = 0 ; i < encrypted.size();i++)
            {
                makeshift(encrypted,i);

                // here is where the error occurs which is "invalid conversion
                // from const char to char
                if(encrypted.find(common) != -1)
                {
                    cout<<"found common"<<endl;
                    cout<<encrypted<<endl;
                    break;
                }
            }
        }
    }
}

Hi CaVeMaN. I've trimmed some of the fat out of your post. Hope that's okay with you. We'll do our best to help and not be too hard on you. :) — John Kugelman, Sep 14 '14 at 04:46
To start, [`while (!message.eof())` **is wrong**](http://stackoverflow.com/questions/5605125/why-is-iostreameof-inside-a-loop-condition-considered-wrong). Check your IO ops rather than assuming they worked. — WhozCraig, Sep 14 '14 at 04:52
And.. your `makeshift` makes no sense whatsoever and invokes **undefined behavior** by dereferencing a improperly obtained pointer. And you need a loop on `common[]`, passing `common[n]` to `find`. You're trying to pass an array of `char const *` to a function that expects a *single* `char const *`. — WhozCraig, Sep 14 '14 at 05:00

John Kugelman · Accepted Answer · 2014-09-14T05:15:29.780

Compile error

First, you say you're getting a compile error on this line:

if(encrypted.find(common) != -1)

The reason for that is because find() expects its argument to be a string, but common is an array of strings. In other words, find() can only search for one word at a time. It can't search for a whole list of words.

To fix that, you'll want to write a loop and check one word each iteration.

Shift function

Next up is the makeshift function. A couple of suggestions there.

There's no real need to call c_str(). You can change encrypted directly by modifying each encrypted[i] in the loop.
It needs to return the result. If you modify encrypted but do not return it then the caller won't see the result.
The printouts ought to be outside the loop. Presumably you just want to see the before and after results after you've decrypted the entire string.

Here's what it looks like with those issues addressed:

string makeshift (string encrypt, int key)
{
    cout << encrypt << endl; // to see what is in file way to debug

    for (int i = 0 ; i<encrypt.size();i++)
    {
        encrypted[i] -= key;
    }

    cout << "decrypted message is " << encryptedc << endl;
    return encrypted;
}

And then you would call it as:

string decrypted = makeshift(encrypted, i);

By the way, I'm not sure if the encrypted[i] -= key; line is completely correct. My guess is you'll need to handle wraparound. Like if you subtract 3 from the letter "A" you should probably wrap around to "X", correct? If so, I'll leave that as a TODO for you.

I/O error checking

Lastly, let's talk about I/O. Specifically, this bit of code:

ifstream message(argv[1]); // get file name typed by user

if(message.is_open()) // check if file is open
{
    while (!message.eof()) // check if we reached end of file
    {
        // get the whole line in file
        getline(message,encrypted);

        ...
    }
}

A good practice in C++ is to check the results of I/O operations (e.g. getline). After you read a line you need to check that the read actually worked. You don't want to continue on if getline() failed (say, because it hit end-of-file).

Conveniently, if you write while (getline(...)) then that'll do a whole bunch of things all at once--it'll check if the file is open, if it's at EOF, and it'll read lines and tell you if the reads were successful. That means you can replace the stuff above with a single loop:

ifstream message(argv[1]); // get file name typed by user

while (getline(message, encrypted))
{
    ...
}

Thank you man, that was great explanation, and I have changed my while to while (getline(message,encrypted) its sad because one of my past professors told me to use it that way, but thanks for this! — , Sep 14 '14 at 05:30

score 0 · Answer 2 · answered Sep 14 '14 at 05:01

You are passing the value of encrypted to makeshift, so whatever you do in makeshift function will not change the encrypted in main function you have 2 solutions here:

you can either pass the pointer of encrypted to the makeshift function

  void makeshift (string* encrypt, int key){
    for (int i = 0 ; i<encrypt.size();i++)
    {   cout<<encrypt->c_str()<<endl; // to see what is in file way to debug
        const char *encryptedc; // this is to convert string to cstring
        encryptedc = encrypt->c_str();
        encryptedc-=key;
        cout<<"decrypted message is "<<encryptedc<<endl;
    }
  }  

  int main (int argc, char* argv[]){
       ...
       makeshift(&encrypted, i);
       ...
  }

Or you can return the value of encryptedc and assign it back to encrypted in main function

  string makeshift (string encrypt, int key){
    for (int i = 0 ; i<encrypt.size();i++)
    {   cout<<encrypt.c_str()<<endl; // to see what is in file way to debug
        const char *encryptedc; // this is to convert string to cstring
        encryptedc = encrypt.c_str();
        encryptedc-=key;
        cout<<"decrypted message is "<<encryptedc<<endl;
        return encryptedc; //you can assign const char* to string
    }
  }  

  int main (int argc, char* argv[]){
       ...
       encrypted = makeshift(encrypted, i);
       ...
  }

Decode a message without a key

2 Answers2

Compile error

Shift function

I/O error checking