Characters to binary... hexadecimal to binary

Basically what the title says... I need to be able to convert a string to binary... then convert that to hexadecimal (cout << hex << stringname; works fine for that part, but I can't save the output), and then the ability to change hexadecimal to binary and convert the binary to the original string... basically code and decode it...

I need the part converted from the characters (it has to have letter support) to binary to be a string so it uses hexadecimal for 0 and 1, not for the letters... so the integer data type wouldn't work.

I'm using Microsoft Visual C++ 6.0 as a compiler, and this is for a console application (DOS).

ANY help would be much appreciated!

Thanks!

Do you mean a string that containes "10001011101" and to convert it into binary?

Maybe you can tokenize the string then use atoi().

Lets say I input the string "programming" into the program. It takes it as a string and puts the value "111 0000" for 'p' and "111 0010" for 'r' and so on.

So I'll end up with a long binary number representing the word "programming"

I don't want to use an insanely long switch statement, so a function would be nice.

I need to convert that to hexadecimal, and store that, too.


Then I need another routine that will convert the hexadecimal back to the binary, and convert the binary to text.


I don't think it is as hard as it sounds, but I don't know how to get the binary values for characters easily.

I'd also need to be able to convert numbers to binary.

THANKS!

for gods sake. computers only understand binary. Your chars are stored as a binary representation anyway. say char is 65 thats 00100001 in 1 byte. You can access the individual bits with c/c++ bitwise operators. This has been dealt with many times before so do a board search.

I did do a search.

And chill out... I know everything is stored in binary, but C++ doesn't have a command that converts it to binary... I know it is STORED in binary, but I need it usable.

I need it to convert NUMBERS and LETTERS.

And how might I convert to HEXADECIMAL and from HEXADECIMAL to BINARY?

Also, where can I learn about XOR encryption? I found an old thread and I need to start from what I already have of this encryption program.

Quote:

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I know everything is stored in binary, but C++ doesn't have a command that converts it to binary

---

What are you trying to say? Is this what you are trying to do?

Code:

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char stuff[] = "abcd";
char* asdf = changetobinary(stuff); //convert "abcd" to binary and store the string version in asdf

---

Quote:

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I need it to convert NUMBERS and LETTERS

---

To what?

Quote:

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And how might I convert to HEXADECIMAL and from HEXADECIMAL to BINARY?

---

If you're talking about converting an int to hexadecimal/binary, then you are a retard, and can shut up right about now. Otherwise, if you are trying to convert a string which represents a hexadecimal number to a string representing a binary number, you can do the research yourself... first, figure out how to convert from hex to bin (and vice versa) and then figure out how to do it digit by digit or something; or, there might be useful functions for you to use that can convert a string in a certain base into an int, and then convert the int into a string using a certain base.

P.S. don't use caps; they don't make people want to help you. Instead, they make people want to smack you over the head with a large trout.

I meant to convert from base 10 to hexadecimal, yes, I know how to do it manually, but I don't want to use a switch statement if it is avoidable.

if the string was "Hello" (this is simplified because I don't want people to know everything about my program, or its useless except as an exersize...) then (yes, this is basically taken from another tutorial, it doesn't have everything I need, though http://www.cpp-home.com/print_t.pl?102)...

H - 72
e - 101
l - 108
l - 108
o - 111

so overall:
72 101 108 108 111

72 - 01001000
101 - 01100101
108 - 01101100
108 - 01101100
111 - 01101111

so...

01001000 01100101 01101100 01101100 01101111

and...

0100100001100101011011000110110001101111

divide that every 6 bits...

010010 000110 010101 101100 011011 000110 1111

and converting it back to base 10:

18 6 21 44 27 6 15

just going to hexadecimal for output is simple... like I said before, cout << hex << varname;

but I need it stored.

Quote:

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divide every six bits

---

:) There's your first problem. Hexadecimal doesn't mean dividing every 6 bits. But anyways, to do what you're trying, this is what I would do:

1) convert each letter of the input into a binary number, and append each such binary number (I mean, its digits) to a std::string.

2) create a tempory char[7] (or 6? Not sure if you need the null 0...).

3) Copy the first six characters of the std::string to the char array.

4) Use atoi() or strtol() or something to convert it into an integer, and store the result in a vector of ints. (no, I don't want to start up the old atoi() vs strtol() war again :( I heard strtol() is better though, but you can pick whatever you choose.)

5) Repeat 3 and 4 until finished.

Of course, you'll have to make sure you don't do an array index out-of-bounds error or anything, when you get to the last batch of characters in the std::string, in case you have fewer than 6 chars to copy to the temporary array.. you can use std::string::size() to check the size of the string, and then use modulus to figure out how many chars you have left at the end of the string, and only copy that many to the temporary array.

Hope this helps!

P.S. This is all theoretical, and might not work - but it's the best I can do right now, since I'm supposed to be doing homework ;)

Quote:

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I know everything is stored in binary, but C++ doesn't have a command that converts it to binary...

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Do you listen man?

Quote:

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You can access the individual bits with c/c++ bitwise operators.

---

lol i dont think that's what he meant in his question... look at his latest post :)

Quote:

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but I need it stored.

---

Ever heard of stringstreams? Very similar to ostreams except they write to memory instead of the screen.

Hunter2, thanks for looking over some of my obvious stupidities and helping me anyway... lol

I meant that that was one layer of encrypting, the way the tutorial had it. I've thought about it for a little while, and the hexadecimal part I had planned IS pointless pretty much.

So really all I need (at the moment, I know how to do the rest) is how to save the text/number string in binary (as a string, not an integer... I need the leading 0s)

I also need to know how to reverse it.

I also need to understand XOR a lot better so that I can use that, too. How would I implement XOR as a function, so that I can run all this stuff different times... so a function to binary and back, etc...

I hope I make sense this time

I need to be able to go from a string like "Hello" to the ASCII code, and from that to binary.

I don't know how to do that without using a very very long switch statement.

I also have 100% no clue at all how to use "bitwise" or what "stringstreams" are


Oh, and one more thing: the 6 bits thing was dividing it up so that it would change the output... simple to do manually... char[x], etc.

PHP Code:

---

void PrintBinary(const char *pszString)
{
    int length = std::strlen(pszString);
    using std::cout;

    for (int position = 0; position < length; position++)
    {
        for (int i = 0; i <= std::UCHAR_MAX; i++)
        {
            cout << pszString[position] & std::pow(2, i);
        }
        cout << std::endl;
    }
}; 


---

Something like that? May need some tweaking, I didn't test it... and i'm extremely tired.

Oh, and the headers you'll need
<climits>
<iostream>
<string>
<cmath>

Thanks.

I think that looks right, but I don't know about part of it... I don't know what using std::cout does

Would that just output it to the screen, or would it somehow save it for later conversions?

I don't know if this would work... but...

PHP Code:

---

void PrintBinary(const char *pszString)
{
    int length = std::strlen(pszString);
    using std::cout;

    for (int position = 0; position < length; position++)
    {
        for (int i = 0; i <= std::UCHAR_MAX; i++)
        {
            pszString[position] = std::pow(2, i);
        }
        cout << std::endl;
    }
}; 


---

Would that change it to binary and store it?

And, if so, how do I reverse that?

No, that won't store it.
using std::cout; says that i'm using the cout object inside namespace std; This statement enables me to use cout << instead of std::cout <<

To store it, you're going to need a vector or something...

PHP Code:

---

std::vector<bool> *PrintBinary(const char *pszString)
{
    int length = std::strlen(pszString);

    std::vector<bool> *binaryStore = new std::vector<bool>[length];

    for (int position = 0; position < length; position++)
    {
        for (int i = 0; i <= std::UCHAR_MAX; i++)
        {
            binaryStore[position]->push_back(pszString[position] & std::pow(2, i));
        }
    }
    return binaryStore;
}; 


---

Remember to #include <vector>
Sorry if this doesn't work.. i'm not too sharp when it comes to templates or the STL. There are many other ways this could be done, too.

I was thinking... why do you even need to "store it" in the first place? Whenever you need the specific bits just use the loop that I gave you. Instead of printing it out, you can do whatever you want with it. It's utterly pointless to store all those values in a vector of bools.

I need it stored because I'm not stopping there... I've got layers upon layers of encryption planned for this program.

No, you don't need to store it. It's already stored. You just need a method of retriving it. I gave one to you.

pszString[position] & std::pow(2, i);

That will tell you whether a particular bit is set or not.

?

I've gotten it working just fine now, only I have one problem. If the binary value has any leading zeros, then it cuts them off because I convert it as an integer. Is there any EASY and SHORT way to add them back in, storing it as a string? I need it to have all 7 bits.

So... if the input is 'a' it typecasts so that it gets the ASCII code for it is 97, so the binary value is 110 0001. That works fine... I get 1100001 for 'a'.

But if the input is '0' for example... ASCII is 48, binary is 011 0000. It works fine but it chopps off the leading zero because it isn't a string... so what is the easiest way to get it to save 0110000 instead of 110000?

I didn't use your function.

At a quick glance (I haven't been tracking this conversation very well), I would say check how many characters are in the result, and if there are fewer than 8, add extra 0's to the front until there are 8 characters.

yeah, I already thought of that but thought there might be a feature. I wanted to do it nice and neat and not a whole bunch of itoa/atoi type things...

Any easy way? I have it stored as a long, but I need it as a string later. I could use something like

Code:

---

if (iBinvalue < 1000000)

---

Right? I still would have the annoying job of shifting the characters over one place in the character array.

This is really weird... I'm trying to put the values in a string manually with a zero first, the iBuffer (the char as binary in a long) works fine if I cout it, but if I output any of the multidimentional array sOutput2 (char sOutput2[1024][7]), it gives me a completely different value... I tried substituting manual changes for itoa, but that doesn't seem to be the problem...

Code:

---

        for (i = 0, iLength = strlen (sOutput); i<iLength; i++)
        {
                iBuffer = to_binary( int(sOutput[i]) );
               
                        if (iBuffer < 1000000)                //check to see if there are 7 digits
                        {
                                sOutput2[i][0] = '0';
                                sOutput2[i][1] = char( (iBuffer%1000000)/100000 );
                                sOutput2[i][2] = char( (iBuffer%100000)/10000 );
                                sOutput2[i][3] = char( (iBuffer%10000)/1000 );
                                sOutput2[i][4] = char( (iBuffer%1000)/100 );
                                sOutput2[i][5] = char( (iBuffer%100)/10 );
                                sOutput2[i][6] = char( iBuffer%10 );
                                sOutput2[i][7] = '\0';
                        }
                        else if (iBuffer >= 1000000)
                        {
                                _itoa(iBuffer,sOutput2[i],7);
                        }

        }



        cout << iBuffer << endl << sOutput2[0] << endl;

---

For example... if I input 'a' into sOutput (not sOutput2), then send it off for binary conversion, the cout line ends up giving me something like

Code:

---

1100001
12231000

---

or if I input '0'...

Code:

---

110000
0:):)

---

It does that if it is done manually or with itoa! And yes, those are actually smileys in the program, not : ).

Code:

---

char *get_binary(const int binaryOutput)
{
        using namespace std;

        char *pszReturn = new char[length];
        pszReturn[length] = '\0';

        for (int i = 0; i < 8; i++)
        {
                pszReturn[i] = (binaryOutput & static_cast<int>(pow(2, i))) ? '1' : '0';
        }
        return pszReturn;
};

---

Would that work for your purposes? Input a character, say 'a'... and itt'l return "10000110".

I've got a working function for the binary part now... but it won't let me convert it to a string.

Also, I have another small error I got while changing part into a function... its at http://cboard.cprogramming.com/showt...threadid=27296 if you're willing to help me there, too.

Basically, the conversion for 'a' works perfectly. It gives me 1100001 just like I want when I cout it. Thats when its stored in the integer iBuffer.

But when I try to convert it to a string in the multi-dimentional array sOutput2 (declared as sOutput2[1024][7]), then cout that, I get 12231000.

Thanks so much for your help so far!

I just realized I need it 8 characters, not 7...

So no matter what it needs a zero before it.
110000 needs to be 00110000 ('0') and 1100001 needs to be 01100001 ('a')

I think your function would work fine if I used strrev, but then I don't learn anything and I'll just have the problem the next time I try to do something like this.

Ack, your code is very difficult to understand. I think you should make a basic function, which takes 1 character, and returns a string containing all of its bits. Try that first. Returning multi-dimensional arrays, and working with multi-dimensional arrays of characters is getting extremely bulky and complicated.

So basically

char *toBinary(char input)
{
// convert input into a string of its bits and return it
}

OK, thats a good idea. I was trying to get it to do each one. the multidimentional array was so that I could store them all in one place and have a much larger number of characters inputted without a thousand strings.

With your code:

PHP Code:

---

char *get_binary(const int binaryOutput)
{
    using namespace std;

    char *pszReturn = new char[length];
    pszReturn[length] = '\0';

    for (int i = 0; i < 8; i++)
    {
        pszReturn[i] = (binaryOutput & static_cast<int>(pow(2, i))) ? '1' : '0';
    }
    return pszReturn;
}; 


---

That doesn't work because "length" is undefined. What do I put there?

How exactly do I use it?

Oops

Code:

---

const int length = 8;
char *get_binary(const int binaryOutput);

int main(int argc, char *argv[])
{
  cout << get_binary('a');
  return 0;
};

---

calling pow() for each bit! hmmmm very efficient!
Ever heard of bitwise operators?
bitwise & can be used to test bits. Read about it in your text books. check previous threads with search facility. Read about stringstreams and how they can help you. Try to write some code and we'll fix it up for you as long as you do your own donkey work.
Oh yeah btw your bit string will be backwards also.

I'll post the compiled version whenever I finish. I've got the binary runnings smoothly! Thanks for your help!

Eibro, how would I reverse your function (i.e. pass binary values to a function and it outputs the characters)

So if I had 011000010110001001100011, it would convert that back to "abc"

Hello. I haven't really been following, but i think i may have an answer. When i did an encryption where i did bit switching, i made three things to help me. I can only remember exactly how i did two of them. I had a struct(?) that had eight single-bit fields, and i had a union between that and a char. The third thing was the function that sent the char to union.ch, and then did the switches on union.bitz.b1, etc. THen again, this may have been JAVA. I blend the two sometimes...

Hope this is of help.

~Inquirer

Your request seemed like an interesting challenge, so I took it a little further than I normally would :)

Code:

---

#include <cmath>
#include <iostream>

struct bitstring
{
        unsigned char bits[8];
};

bitstring get_binary(const char code)
{
        bitstring returnstring;
        memset(reinterpret_cast<void *>(&returnstring), '\0', sizeof(returnstring));

        int count = 0;
        for (int bit = 128; bit >= 1; bit >>= 1)
        {
                returnstring.bits[count] = (code & bit) ? '1' : '0';
                count++;
        }
        return returnstring;
};


char get_character(const bitstring instring)
{
        char cReturn = '\0';

        int count = 0;
        for (int bit = 128; bit >= 1; bit >>= 1)
        {
                        if (instring.bits[count] == '1')
                        {
                                cReturn |= bit;
                        }
                        count++;
        }
        return cReturn;
};

int main(int argc, char* argv[])
{
        using namespace std;

        bitstring output = get_binary('1');
        char input = get_character(output);
        cout << input;

        return 0;
}

---

Oh, and Stoned_Coder, i've dropped the pow... as you suggested.

Eibro, could you explain how this works? Especially the part that says ...<void *>... ?

And, i had to tweak it as follows so it would display correctly.

Code:

---

#include <cmath>
#include <iostream>

struct bitstring
{
        unsigned char bits[9]; // I made this to leave room for '\0'
};

bitstring get_binary(const char code)
{
        bitstring returnstring;
        memset(reinterpret_cast<void *>(&returnstring), '\0', sizeof(returnstring));

        int count = 0;
        for (int bit = 128; bit >= 1; bit >>= 1)
        {
                returnstring.bits[count] = (code & bit) ? '1' : '0';
                count++;
        }
  returnstring.bits[8]='\0'; // I added this
        return returnstring;
};


char get_character(const bitstring instring)
{
        char cReturn = '\0';

        int count = 0;
        for (int bit = 128; bit >= 1; bit >>= 1)
        {
                        if (instring.bits[count] == '1')
                        {
                                cReturn |= bit;
                        }
                        count++;
        }
        return cReturn;
};

void binary()
{
        using namespace std;

        bitstring output = get_binary((char)getche());
        char input = get_character(output);
        cout << "\"" << input << "\"" << endl;
  cout << output.bits << endl;
  if (getche());

        return;
}

---

memset(reinterpret_cast<void *>(&returnstring), '\0', sizeof(returnstring));

reinterpret_cast<void *>(&returnstring) is using C++ style casting to cast a void pointer on the address of returnstring. Using C-Style casting it'd be (void *)&returnstring

Also, changing the number of bits to 9 was unnessicary, i'm only using 8 characters (0-7). The 8th contains the null terminator.

128, 64, 32, 16, 8, 4, 2, 1... There's only 8 values there, these are the ones i'm using.

Anything else need clarification?

Uhh... yeah :D What's the advantage of using c++-casting instead of C casting?

ummm... yes...

1) why exactly would you WANT to cast something into a void pointer?

2) About the bits[9], if you output it to the screen, it also outputs he next few variables because when you say bits[8] you are setting aside memory for bits[0] to bits[7], not bits[0] to bits[8].
the null terminator gets put in the last one later, and all 8 bits are outputted to the screen.

3) What does memset do? I haven't done much memory stuff...

4) What is the memset prototype?

5) I don't exactly see how the loop(s) work to translate them to and from the binary... I get how you are stepping down the 128...1 ladder, but how do you use that to calculate whether the bit was on or off??? ::confused::
[EDIT]
I have figured out #5... except for the following:

Code:

---

returnstring.bits[count] = (code & bit) ? '1' : '0';

---

I can see that this would only leave the bit in question set, but i don't see how this would test true. Does null evaluate to false? That must be it...


Thanks for the example! I learned a lot!
[/EDIT]

Thanks...
~Inquirer

Anyone who knows JAVA should help on this one...

Is it java or c++ (or both) that have unions?
Is it java or c++ (or both) that have single-bit data members?
[DELETED][/DELETED]

~Inquirer

Both might have unions (I haven't touched java for a while), but I'm pretty sure only c++ has single-bit data members.

And about casting to a void pointer, that's 'cuz memset takes a void pointer (well, I'm guessing that), and some compilers won't do the typecast automatically. But I still don't see why he didn't just use (void*)...

memset(sometype *pointerToDataStructureOrVariableButProbablyDataSt ructure, char valueToSetTo, int howManyBytesToSet)

-sometype is probably either void or char - i'm guessing void, 'cuz otherwise eibros wouldn't have typecasted the thing to void.
-valueToSetTo might be int or unsigned char.
-howManyBytesToSet may be a long or something.
-valueToSetTo is usually set to 0.
-howManyBytesToSet is usually filled in with sizeof(the datastructure).
-memset is usually used to clear out a datastructure.
-another way to clear out a datastructure would be to use ZeroMemory(), which is either a macro or inline function or something that uses memset() and just fills in valueToSetTo with 0.

Yes, it's typecasted to void * because that's what memset takes. C++ allows for many different casts: dynamic_cast<> static_cast<> reinterpret_cast<> const_cast<>, I'm guessing these employ better typesafty and more efficient casting than the regular C-Style casting. (someone who knows more about this, please, please post)

About bits[9], I still don't see how i'm using more than 8 elements. 0-7 is all i'm using. You're correct in adding bits[8] = '\0' though. When I declare bits[8] I get 9 elements- 0-7 are useable to me, and #8 is the null terminator. "h, e, e, l, l, l, o, o, '\0'"

memset(void *, int, size_t); is the prototype for memset.

Code:

---

returnstring.bits[count] = (code & bit) ? '1' : '0';

---

We're taking 'bit' (which is 128, 64, 32, 16, 8, 4, 2, or 1... binary digits) and ANDing with 'code'. If both 'code' and 'bit' have the particular bit set, then it returns true. If 'code' doesn't have the bit set, it returns false.

Take the letter 'a': 01100001
Now, throughout the loop, bit is set to 00000001, 00000010, 00000100, 00001000, 00010000, 00100000, 01000000.
Throughout the loop, the entries bolded are the ones which return true.

Well i'm stumped...
Well, I was stumped. (Those two lines were written a half hour apart)

I'm wondering why output's bits weren't retaining the value returned from the function...
Something to do with it going out of scope? Shouldn't the object be copied?

Oh well, I came up with this:

Code:

---

// ****msvc.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"

#include <cmath>
#include <iostream>
#include <assert.h>

class bitstring
{
public:

        bitstring();
        bitstring(const bitstring &bs);
        ~bitstring();

        unsigned char *getbits();
        void setbit(int bitnum, int val);
        unsigned char getbit(int bitnum);       

        void setbitstring(const char code);
        char getcharacter();

private:
        unsigned char *bits;
};

bitstring::bitstring()
{
        bits = new unsigned char[8];
        memset(reinterpret_cast<void *>(bits), 8, sizeof(bits));
}


bitstring::~bitstring()
{
        delete[] bits;
}

bitstring::bitstring(const bitstring &bs)
{
        bits = new unsigned char[8];

        for (int i = 0; i < 8; i++)
        {
                bits[i] = bs.bits[i];
        }
        bits[8] = '\0';
}


unsigned char *bitstring::getbits()
{
        return bits;
}

void bitstring::setbit(int bitnum, int val)
{
        assert(bitnum < 8);
        bits[bitnum] = val;
}

unsigned char bitstring::getbit(int bitnum)
{
        assert(bitnum < 8);
        return bits[bitnum];
}

char bitstring::getcharacter()
{
        char cReturn = '\0';

        int count = 0;
        for (int bit = 128; bit >= 1; bit >>= 1)
        {
                        if (getbit(count) == '1')
                        {
                                cReturn |= bit;
                        }
                        count++;
        }
        return cReturn;
};

void bitstring::setbitstring(const char code)
{
        int count = 0;
        for (int bit = 128; bit >= 1; bit >>= 1)
        {
                setbit(count, (code & bit) ? '1' : '0');
                count++;
        }
        bits[8] = '\0';
};

int main(int argc, char* argv[])
{
        using namespace std;

        bitstring output;
        output.setbitstring('a');

        cout << output.getbits() << endl;

        cin.get();
        return 0;
};

---

Looks like I went a little overboard...

lol... let me try writing a simple function.

I did some research, and only c++ has unions and single-bit data structures. Now, through the demonic ingenuity of my thought processes, i recalled how i did the bit access.

Code:

---

union char_bin_hex {  //edit: changed struct to union
  char ch;
  union char_to_bin bits;
  union char_to_hex dibits;
}

struct char_to_bin {  //edit:  changed union to struct
  unsigned b1: 1
  unsigned b2: 1
  unsigned b3: 1
  unsigned b4: 1
  unsigned b5: 1
  unsigned b6: 1
  unsigned b7: 1
  unsigned b8: 1
}

struct char_to_hex {  //edit: changed union to struct
  unsigned d1: 2
  unsigned d2: 2
  unsigned d3: 2
  unsigned d4: 2
}

---

All you do is put the char into char_bin_hex.ch and access the individual bits/dibits through char_bin_hex.bits.b1 etc and char_bin_hex.dibits.d1 etc.

Hope this helps! I will work on a function tht works with this when iget back frm church & lunch.

~Inquirer

Quote:

---

Originally posted by Inquirer
I did some research, and only c++ has unions and single-bit data structures. Now, through the demonic ingenuity of my thought processes, i recalled how i did the bit access.

Code:

---

struct char_bin_hex {
  char ch;
  union char_to_bin bits;
  union char_to_hex dibits;
}

union char_to_bin {
  unsigned b1: 1
  unsigned b2: 1
  unsigned b3: 1
  unsigned b4: 1
  unsigned b5: 1
  unsigned b6: 1
  unsigned b7: 1
  unsigned b8: 1
}

union char_to_hex {
  unsigned d1: 2
  unsigned d2: 2
  unsigned d3: 2
  unsigned d4: 2
}

---

All you do is put the char into char_bin_hex.ch and access the individual bits/dibits through char_bin_hex.bits.b1 etc and char_bin_hex.dibits.d1 etc.

Hope this helps! I will work on a function tht works with this when iget back frm church & lunch.

~Inquirer

---

Is it just me or did you get that a little backwards? The unions should be structures, and the structure should be a union. Or else i'm an idiot. (Then again i've never done anything like that before)

Yeah! You're right!

I am editing that now, so don't get confused by eibro's post guys...

Now i can actually make the function correctly!

::: see my new thread :::