Please help me with a code to read the value of a byte in a file.

Hello

I'm new to c++ and to all forms of programming and am trying to get the source code for a program that prints out the hex value of any given byte in a file.

I've gone through; this site, google, yahoo, cplusplus.com, C++ for dummies 5th Edition and, unbelievably, more. The answer just seems to elude me. No one says how to do this. I can only get the size of a file or read a txt file, but not the hex, or any other value of any given byte in a file. Thats the problem. Can ANYONE show me the source of a simple program that does this?

I'm going to show you the closest I could get to this a majority may lose your lunch when you see it but you'll understand what a state I'm in with which I can only get the character of in a txt file and not the value of a byte of anyfile. I know I'm supposed to make the char unsigned but that stops me from compiling at all and give me errors.

[tag]

Code:

---

#include <iostream>
#include <fstream>
using namespace std;
int x, y;
int main () {
char * buffer;
  Start:
  cout << "Enter the number of the byte you are looking for: ";
  cin >> x;
  if (x<1) {
  cout << "We begin counting at 0. Please try again.\n"; goto Start;}
  ifstream is;
  is.open ("Yo.bin", ios::in|ios::binary);
  is.seekg (x, ios::beg);
  is.read (buffer,1);
  is.close();
  cout << "\nThe hex value of byte number"<< x <<" is ";
  cout.write (buffer,1);
  cout << "\n\nDo you want to return to the start of this program?\n";
  cout << "Press 1 for YES or any other key for NO: ";
  cin >> y;
  if (y == 1) {goto Start;}
  if (y == !1) {goto End;}
  End:
  cout << "Goodbye!\n";
  delete[] buffer;
        return 0;}
[/tag]

---

Thanks for reading.

Whether something is displayed as a hex value or a decimal value or a character is just a matter of output representation. The byte itself is still the same.
To change the representation you can use I/O format flags. For example:

Code:

---

#include <iostream>
int main()
{
    char c = 'A';
    std::cout << std::showbase << std::hex << int(c);
}

---

In your code, buffer doesn't seem to be allocated, and since you are going to put just one character into it, may-be it could be declared as

Code:

---

char buffer[1] = {0};

---

(May-be you could also read into a single char, but that would probably require ugly casts.)

You never allocate memory for buffer (and yet you delete it).

Code:

---

char *buffer = new char[2];

---

Or, seeing as you're only reading a single char anyway,

Code:

---

char buffer;

---

and get rid of the delete call.

Gotos? In *my* C++ program?

Code:

---

int y = 0;
while (!y)
{
    // Do your thing
}

---

Is this a "find five problems"?

You don't need that much code to make me loose my lunch, surely this can easily be written without goto's:

Code:

---

  if (x<1) {
  cout << "We begin counting at 0. Please try again.\n"; goto Start;}

---

Note that I'm not religiously against goto's, just think they should be avoided when there is a better, simpler alternative.
Also, your error message doesn't seem to match exactly with the if-statement.

Second problem:

Code:

---

char * buffer;

---

This is a pointer to memory, but it's never assigned to any particular memory location, so it's going to crash, I suspect. Use a char variable, and pass the address to read.

Third problem:

Code:

---

cout.write (buffer,1);

---

This will display the ascii character (or whatever the local font representation of the data is), not the hex-value. To show a hex value, you need something like

Code:

---

cout << hex << buffer;

---

Fourth problem:

Code:

---

  cout << "Press 1 for YES or any other key for NO: ";
  cin >> y;

---

This would fail if the user types in "a" for an answer.

Problem five:

Code:

---

  if (y == !1) {goto End;}
  End:

---

So, not only do you use goto for ABSOLUTELY no reason, your if-statement is also not doing what you expect - not that you'd notice, because you end up at exactly the same place whether the if-statement is true or not, as there is no code between the goto and the lable end, but I challenge you to find a way to NOT print the message in this code:

Code:

---

  if (y == !1) {goto End;}
    cout << "Got here" << endl;
  End:

---

--
Mats

> delete[] buffer;
You didn't allocate it, so you can't use it and you can't delete it.

Also, use loop constructs in place of those goto statements.

Also, there is no point in x and y being global.

Edit: Meh, beaten *3

Quote:

---

Originally Posted by Salem

> delete[] buffer;
You didn't allocate it, so you can't use it and you can't delete it.

Also, use loop constructs in place of those goto statements.

Also, there is no point in x and y being global.

Edit: Meh, beaten *3

---

And I just realized it wasn't find five faults, but find eight faults... ;)

--
Mats

you had the right idea, but a few details were off. consider the following:

Code:

---

        int x;
        bool go=true;
        while(go)
        {
                cout << "Enter the number of the byte you are looking for: ";
                cin >> x;
                if (x<0)
                {
                        cout << "We begin counting at 0. Please try again.\n";
                }
                else
                {
                        unsigned int y;
                        char buffer;
                        ifstream is;
                        is.open ("Yo.bin", ios::binary);
                        is.seekg (x, ios::beg);
                        is.read (&buffer,1);
                        is.close();
                        cout << "\nThe hex value of byte number "<< x <<" is " << hex << (int)buffer << endl;
                        cout << "\n\nDo you want to return to the start of this program?\n";
                        cout << "Press 0 for NO or any other key for YES: ";
                        cin >> y;
                        if(!y)
                        {
                                go = false;
                        }
                }
        }
        cout << "Goodbye!\n";
        return 0;

---

good luck.

Code:

---

cout << "\nThe hex value of byte number "<< x <<" is " << hex << (int)(unsigned char)buffer << endl;

---

if you want to avoid having it sign-converted.

--
Mats

could you not just cast it once to unsigned int?

Quote:

---

Originally Posted by m37h0d

could you not just cast it once to unsigned int?

---

Perhaps, but I feel that it will still sign-extend to int first, then "unsign" that.

--
Mats

I wanted to come back to thank everyone on this thread for their help, especially m37h0d for putting it in terms the layman to the world of c++ can understand.

Thanks m37h0d!

http://cpwiki.sourceforge.net/Common...llocating_them

Quote:

---

Originally Posted by koxson

I wanted to come back to thank everyone on this thread for their help, especially m37h0d for putting it in terms the layman to the world of c++ can understand.

Thanks m37h0d!

---

i did?

yay me!

making the code as clean and simple as possible should always be a goal imo. i didn't write that in any fashion that was supposed to be "for the layman"; i just tried to make it as clear and simple as possible, which as far as i'm concerned is always good programming!