Ignoring input.

I have been making simple programs to solidify my basic knowledge of C++. But while making this Binary converter I came across a problem.

I cant seem to figure out how to ignore input after 8 characters. I want to read in the first 8 characters the user inputs and discard the rest.

If cin.ignore is an option I cant figure out how to do it.

Code:

---

#include <iostream>

int BinToDec ( char *bits )
{
       
        int result = 0;
        int mask = 1;
       
        for ( int loop = 7; loop >= 0; loop -- ) {
               
                if ( bits [loop] != '0' && bits [loop] != '1' ) {
                       
                        std::cout << "Error: Invalid Input\n";
                        return -1;
                       
                }
               
                else if ( bits [loop] == '1' ) result += mask;
               
                mask *= 2;
               
        }
       
        std::cout << "\nResults:\n";
        std::cout << "Bin: " << bits << '\n';
        std::cout << "Dec: " << result << '\n';
        std::cout << "Ascii: " << static_cast <char> ( result ) << '\n';
       
        return 0;
       
}

int main ()
{
       
        char bit_array [8];
        memset ( bit_array, 0, 8);
       
        std::cout << "You must enter 8 bits. Q to exit\n\n";
       
        while ( 1 ) {
               
                std::cout << "\nEnter a binary value: ";
                std::cin  >> bit_array;
               
                if ( bit_array[0] == 'q' ) break;
               
                BinToDec ( bit_array );
               
        }
       
        std::cout << "\nGoodbye.\n";
       
        return 0;
       
}

---

:confused:

cin.get( &bit_array[0], 8 );

>>&bit_array[0]
bit_array

**EDIT**
Also, consider
cin.read(bit_array, 8);

I'm not 100% sure that'll work properly on cin (don't see why not though), but it's what I do when I want to read (n) bytes from a file.

Mm.

Tried both of those and neither one worked. It had very odd behavior. It would work different everytime i tried it. Sometimes it would completely ignore everything i put in. ??

Hi Vicious,
did you consider using get() method?
I would suggest something like this:

Code:

---

int main ()
{
       
        char bit_array [9];
        memset ( bit_array, 0, 9);
       
        std::cout << "You must enter 8 bits. Q to exit\n\n";
       
        while ( 1 ) {
               
                std::cout << "\nEnter a binary value: ";
               
                std::cin.get(bit_array,9);

                if(std::cin.get() !='\n')
                {
                        std::cin.clear();
                        while((std::cin.get() !='\n'));
                }
               
                if ( bit_array[0] == 'q' ) break;
               
                BinToDec ( bit_array );       
        }
        std::cout << "\nGoodbye.\n";
        return 0;
}

---

I didn't test all possibilities but think it's OK.
Cheers!

Vicious, what's your code (full listing preferred, if it's short enough)?

There are plenty of ways to read the eight (?) characters that you want. Discarding everything after that is equally varied. Two fairly common ones that don't use an explicit loop are:

Code:

---

#include <limits>

cin.ignore ( numeric_limits<streamsize>::max(), '\n' );

---

and

Code:

---

cin.ignore ( cin.rdbuf()->in_avail() );

---

A quick implementation is obvious once you have those:

Code:

---

#include <iostream>
#include <limits>

using namespace std;

int bit_convert ( char bits[] )
{
  int dec = 0;
  int base = 1;

  for ( int i = 7; i >= 0; i-- ) {
    if ( bits[i] != '0' )
      dec += base;

    base <<= 1;
  }

  return dec;
}

int main()
{
  char bits[9];

  cout<<"Enter an octet: ";
  if ( cin.getline ( bits, sizeof bits ) ) {
    cin.ignore ( cin.rdbuf()->in_avail() );

    cout<<"The value of "<< bits <<" is "<< bit_convert ( bits ) <<endl;
  }
}

---

Slightly harder is handling errors and realizing that a byte isn't always eight bits, then changing your code to deal with that. :)

Quote:

---

Originally Posted by Hunter2

Vicious, what's your code (full listing preferred, if it's short enough)?

---

Heh thats the full code in the first post.


And thank you Prelude. I do not believe I have ever used these methods this way. :rolleyes:

>>Heh thats the full code in the first post.
Oh, I thought maybe you updated it when playing around with get() and read().

>>And thank you Prelude. I do not believe I have ever used these methods this way.
Ditto.

Quote:

---

Originally Posted by Prelude

Slightly harder is handling errors and realizing that a byte isn't always eight bits, then changing your code to deal with that. :)

---

Can you give some examples where byte isn't always eight bits?

Quote:

---

Originally Posted by Micko

Can you give some examples where byte isn't always eight bits?

---

Do a search...

>Can you give some examples where byte isn't always eight bits?
The only machines I know of are old (like the PDP, Honeywell and Cray before they started allowing 8-bit addressable units). I believe, though I'm not sure, that some modern mainframes have variable byte sizes, such as 7, 8, or 9-bit bytes where 9 is the most common, and probably a few embedded architectures as well.

I work with PCs, but even there it's not wise to assume that a byte is 8 bits or that it will remain that way if it is. :) The story about Microsoft programmers going through every line of their C compiler's source code because they assumed an int was always going to be 16 bits and the designers had other plans.