decimal to binary conversion

Hello,

The next exercide I have is to convert a decimal to a binary.
So I made this :

Code:

---

      int
main ( int argc, char *argv[] )
{
        int nummer, uitkomst, remainder ;
        char uitkomst2 ;
        nummer = 4 ;
        uitkomst = 4 ;
        while (uitkomst <0)
        {
                uitkomst = nummer/2;
                remainder = nummer%2;
                if (remainder == 1)
                        {
                                uitkomst2 = "1" + uitkomst2 ;
                        }
                else
                        {
                                uitkomst2 = "0" + uitkomst2;
                        }
                nummer = uitkomst;
        }
        printf ("uitkomst is %s", uitkomst2);
        return EXIT_SUCCESS;
}                              /* ----------  end of function main  ---------- */

---

It won't compile because uitkomst2 = "0/1" + uitkomst2 is wrong.
I get this error :
test2.c|34 col 15| warning: assignment makes integer from pointer without a cast

What I'm trying to achieve is this:

Let says the number is 4
Then 4/2 = 2 with no remainder so it will be a 0
Then 2/2 = 1 with no remainder so it will be a 0
Then 1/2 = 0.5 a remainder so it will be a 1
So uitkomst will be 001

Oke, almost right.

But what Am i thinking wrong here.? I know I have to read the outcome from end to beginning.

Roelof

Well... you actually have the correct answer... but you have it backwards... which is an artifact of displaying the number as you decode it...

Think about it... you are breaking the number down right to left but printing left to right...

Try to think of a way to burst the number from left to right... (since you can't reverse the printing) ... It's not hard to do, but it does require a different approach.

Why are you trying to add the string value "1" to a single character?

Quote:

---

Originally Posted by WaltP

Why are you trying to add the string value "1" to a single character?

---

Because he's still learning... Your observation is correct and it should be fixed... but first he has to work out the process.

Hello,

I did that because i work backwards,
So I thought if I take the old piece and put a 1 or 0 before it I don't have to revert the printing.

Only thing that comes in mind is to divide till the it's 0.
Then you have the first number and work backwards.
But then you talk about recursion and that's not discussed in the book yet.

So still thinking and use google.

Roelof

Edit 1 :

I have found this page : How to Convert from Decimal to Binary - wikiHow
And I can use this method : Comparison with descending powers of two and subtraction
But the downside of this method is that if a user entertered a big number I need a lot of substractions.
So how big must the last number be.

Hello

Anyone who can shine a light on this.

With the method1 (which Im using now) I have no idea how to store the 0 and 1 on the right place because you can't know how many 0 en 1 you need. You can store in a a array and read it from the last to the first but you don't know which number the lastest entry has.

Method2 jhas the disadvantage that you first has to know which power of 2 is the nearest.

Roelof

Quote:

---

Originally Posted by roelof

With the method1 (which Im using now) I have no idea how to store the 0 and 1 on the right place because you can't know how many 0 en 1 you need. You can store in a a array and read it from the last to the first but you don't know which number the lastest entry has.
Roelof

---

Except, of course, you do. How else do you know where to put the element in the array when you compute it, if you don't know where the end of the array is?

Oke

of course I can use let's say a array of 20.
And every time I divide it with 2 a sort of counter can be used to put a array in it
So what i mean it array[place++] = remainder.

It that what you mean or do we misunderstood each other ?

Roelof

Quote:

---

Originally Posted by roelof

Oke

of course I can use let's say a array of 20.
And every time I divide it with 2 a sort of counter can be used to put a array in it
So what i mean it array[place++] = remainder.

It that what you mean or do we misunderstood each other ?

Roelof

---

That's what we mean. And when you get to the end you can start at place (well, really, place-1 the way you have it written) to go backwards.

Oke,

I see one problem with this method.
We cannot know how big the array must be,
If the user enters a really big number I need a big array for all the 1 and 0.

Roelof

It doesn't matter what the user types in; even if the user types in 99999999999999999999999999999999999999999999999999 99999999999999999 the computer will not store a number larger than the data type can handle.

Oke,

But that means if the user enters that number the outcome will not be bigger then 20 so the outcome is not right.

I think if the divine reached the 21 number and try to store it it gives a error because you are outside the boundarys of that array.

Or im mistaken here ?

Roelof

Quote:

---

Originally Posted by roelof

Oke,

I see one problem with this method.
We cannot know how big the array must be,
If the user enters a really big number I need a big array for all the 1 and 0.

Roelof

---

Sure we do... 32 for 32 bit binary, 64 for 64 bit binary, etc... Then you limit the user's input to fit the array.

But there is a simpler way still.... (although your course may not have covered bitwise functions, yet)

For a simple 32 bit signed integer containing a positive value...

Code:

---

int number = 112323; // for example
int filter = 0x40000000u;  // msb = 1;

while (filter)
  { if (number & filter)
      printf("1");
    else
      printf("0");
    filter = filter >> 1; }
printf("\n\n");

---

Oke,

I understand that,
But how can I know if a number is a 32 bit binary or a 64 bit binary.
In my opinion we can know that after the conversion.

Roelof

Quote:

---

Originally Posted by roelof

Oke,

I understand that,
But how can I know if a number is a 32 bit binary or a 64 bit binary.
In my opinion we can know that after the conversion.

Roelof

---

That depends on the hardware on your machine, and how your environment talks to the hardware. (In other words: when you type "int" you will get either a 32-bit or a 64-bit integer (or I suppose a 16-bit integer in the worst case), depending on your particular system. You will always get the same-sized integer every time, until you do something like buy a new system or a new compiler.)

Quote:

---

Originally Posted by roelof

Oke,

I understand that,
But how can I know if a number is a 32 bit binary or a 64 bit binary.
In my opinion we can know that after the conversion.

Roelof

---

You know that by knowing about your compiler... in my compiler int is 32 bits... I know this and I can trust it.
If your compiler's library includes stdint.h you can be very explicit about variable sizes... for example int32_t guarantees a 32 bit integer.

If you are concerned about the user's entry overflowing the integer's range... you need to limit your user's input to acceptable values (almost always a necessity in real world programming, anyway)... There are a great many ways of doing this but in a training exercise it's safe to make a note to work on this later (once your more familiar with the language)

Oke,

So I can't know if I have to have to array of 32 of 64 bit.

Roelof

Quote:

---

Originally Posted by roelof

Oke,

So I can't know if I have to have to array of 32 of 64 bit.

Roelof

---

Again... you know this because you are telling the compiler what to do... there's no way to back-track and discover this later. The C compiler won't track this for you and it won't produce errors if you mess up. It is up to you as a programmer to keep track of the variables and their sizes.

Here's a little example of limiting input to acceptable values, based on my previous example... (Note this is untested but it should be pretty close...)

Code:

---

#include <stdint.h>
#include <stdio.h>

int main (void)
  { int64_t number;  // 64 bit input space
    int32_t  filter = 0x40000000u;  // 32bits, msb = 1;

    while (number > -1)
      { printf("Enter a number from 0 to %d  (-1 to exit) :", INT32_MAX);

        // read in a 64 bit value
        scanf("%lld", number);

        // get rid of extraneous newline character
        getchar();

        // range check
        if ((number > -1) && (number <= INT32_MAX)
          { printf("\n\n");
              break; } }

      // exit on any negative number
      if (number < 0)
        exit(0);

      printf("Your number as binary is :  ");

      // burst to binary
      while (filter)
        { if (number & filter)
            printf("1");
          else
            printf("0");
          filter = filter >> 1; }

      printf("\n\n");

  return 0; }

---

Oke,

I hope my book (C book) will explain that later.
It's not something chapter 1 has explained.
They only talked about char and int.

But thanks for the explanation.

Roelof

Quote:

---

Originally Posted by roelof

Oke,

So I can't know if I have to have to array of 32 of 64 bit.

Roelof

---

The C standard requires your compiler to tell you which it is in the manual/other documentation that came with the compiler. So assuming you know where you put the manual/other documentation, there will be a section about "implementation limits" or some such that says whether "int" is 32 or 64 bits.

Oke,

And if I understand it right
It can be that on my system int us 64 bit and on another system int = 32 bit.
So as Commentator have said it better that I say if a int is 32 or 64 bit.
Otherwise wierd things can happen.

Roelof

Quote:

---

Originally Posted by roelof

Oke,

I hope my book (C book) will explain that later.
It's not something chapter 1 has explained.
They only talked about char and int.

But thanks for the explanation.

Roelof

---

No worries... In your place I'd make a note to check into it a bit later in the learning process... When I was learning C++ (a recent undertaking) I made lots of notes about stuff I have to review and investigate later. It's all part of wanting to learn....

Quote:

---

Originally Posted by roelof

Oke,

And if I understand it right
It can be that on my system int us 64 bit and on another system int = 32 bit.
So as Commentator have said it better that I say if a int is 32 or 64 bit.
Otherwise wierd things can happen.

Roelof

---

It doesn't always matter... what do you care when simply counting to 10? But when doing binary manipulations or dealing with really big numbers, yes, it is better.

FWIW... there are some cases where an int = 16 bits, too.

Although I doubt your textbook has covered it yet, you can always do this:

Code:

---

int x = sizeof(int):

---

The value of x is in characters (bytes) so...

2 = 16 bit integers
4 = 32 bit integers
8 = 64 bit integers

Oke,

I will try to make the exercise and post the code here for checking.
Both thanks for the help and patience.

Roelof

Hello,

I need your help one more time.
I now have this code:

Code:

---

/*
 * =====================================================================================
 *
 *      Filename:  test2.c
 *
 *    Description:  conventer from dec to 0101
 *
 *        Version:  1.0
 *        Created:  14-05-11 10:03:25
 *      Revision:  none
 *      Compiler:  gcc
 *
 *        Author:  Dr. Fritz Mehner (mn), [email protected]
 *        Company:  FH Südwestfalen, Iserlohn
 *
 * =====================================================================================
 */

#include        <stdio.h>
#include        <stdlib.h>
        int
main ( int argc, char *argv[] )
{
        int nummer, uitkomst, teller, remainder ;
        char uitkomst2[10] ;
        nummer = 4 ;
        uitkomst = 4 ;
        while (uitkomst <0)
        {
                uitkomst = nummer/2;
                printf ("uitkomst : %d", uitkomst);
                remainder = nummer%2;
                uitkomst2[teller]= remainder;
                teller = teller +1 ;
                nummer = uitkomst;
                printf ("teller:%d  nummer:%d",teller, nummer);
        }
        while (teller !=0)
        {
                printf("%d", uitkomst2[teller]);
                teller = teller -1 ;
        }

        return EXIT_SUCCESS;
}                              /* ----------  end of function main  ---------- */

---

I now get no output except a cryptic command terminated message.

Anyone see what I do wrong here ?

Roelof

It's making me a little dizzy the way you're flipping numbers around in there...
For example...

Code:

---

remainder = nummer%2;
 uitkomst2[teller]= remainder;
 teller = teller +1 ;

---

Can be combined as...

Code:

---

uitkomst2[teller++]= nummer%2;

---

with exactly the same result.

It's always a good idea to go through and minmize your code, eliminating any unnecessary variables and combining statements when you can.

Oke,

I changed the code to this :

Code:

---

/*
 *  * =====================================================================================
 * 
 *          Filename:  test2.c
 *   
 *          Description:  conventer from dec to 0101
 *     
 *          Version:  1.0
 *          Created:  14-05-11 10:03:25
 *          Revision:  none
 *          Compiler:  gcc
 *           
 *          Author:  Dr. Fritz Mehner (mn), [email protected]
 *          Company:  FH Südwestfalen, Iserlohn
 *             
 * =====================================================================================
  */

#include        <stdio.h>
#include        <stdlib.h>

int
main ( int argc, char *argv[] )
        {
            int nummer, uitkomst, teller ;
            char uitkomst2[10];
            nummer = 4 ;
            uitkomst = 4 ;
            while (uitkomst <1)
                {
                        uitkomst = nummer/2;
                        uitkomst2[teller++]= nummer%2;
                        printf ("teller:%d  nummer:%d",teller, nummer);
                }
                while (teller !=0)
                  {
                        printf("%d", uitkomst2[teller]);
                        teller = teller -1 ;
                }
  return EXIT_SUCCESS;
        }                              /* ----------  end of function main  ---------- */

---

But still no output and a cryptic message command terminated.

Anyone a idea waht happens.

Roelof

Oke,

I found it.
I forget to set teller to 0

But the problem of no output is still here

Anyone who has tips for that.

Roelof

Three guesses as to how many times this loop runs:

Code:

---

            uitkomst = 4 ;
            while (uitkomst <1)
                {
                        uitkomst = nummer/2;
                        uitkomst2[teller++]= nummer%2;
                        printf ("teller:%d  nummer:%d",teller, nummer);
                }

---

Quote:

---

Originally Posted by roelof

Oke,

I found it.
I forget to set teller to 0

But the problem of no output is still here

Anyone who has tips for that.

Roelof

---

Ok... look at the values of each variable as you come into the while loops...

nummer = 4, uitkomst = 4, teller = 0;
then you enter the loop... guarded by while (uitkomst less than 1)
uitkomst = 4
the loop never executes.
Now it falls through to the second loop... while (teller != 0)
Teller was unchanged by the first loop so teller = 0;
The second loop never executes.

Hense... no output.

Hello,

I saw it.
I have now this

Code:

---

/*
 *  * =====================================================================================
 * 
 *          Filename:  test2.c
 *   
 *          Description:  conventer from dec to 0101
 *     
 *          Version:  1.0
 *          Created:  14-05-11 10:03:25
 *          Revision:  none
 *          Compiler:  gcc
 *           
 *          Author:  Dr. Fritz Mehner (mn), [email protected]
 *          Company:  FH Südwestfalen, Iserlohn
 *             
 * =====================================================================================
  */

#include        <stdio.h>
#include        <stdlib.h>

int
main ( int argc, char *argv[] )
        {
            int nummer, uitkomst, teller, teller2;
            char uitkomst2[10];
            nummer = 4 ;
            uitkomst = 4 ;
            teller = 0 ;
            while (uitkomst > 0)
                {
                        uitkomst = nummer / 2 ;
                        uitkomst2[teller++]= uitkomst%2;
                        nummer = uitkomst;
                        printf ("teller:%d  deling:%d uitkomst deling:%d  \n",teller, uitkomst%2, uitkomst);
                }
            teller2 = 1 ;
          while (teller2 < teller)
                  {
                        printf("%d", uitkomst2[teller2]);
                        teller2 = teller2 + 1 ;
                  }
  return EXIT_SUCCESS;
        }
/*c----------  end of function main  ---------- */

---

It working now only not good.
4 gets now the answer 10 where it must be 100

Roelof

That's because you divide by 2 before trying to convert the number (as the first line in your while loop). Don't do that.

Quote:

---

Originally Posted by tabstop

That's because you divide by 2 before trying to convert the number (as the first line in your while loop). Don't do that.

---

Old joke....

Quote:

---

A guy walks into the doctor's office and hooks his thumb over the edge of the desk.
He pulls out a claw hammer and whacks himself right on the thumb.

"Doc, it hurts like heck every time I do that"

The doctor replied: "Then don't do that!".

---

Oke,

I changed it to this:

Code:

---

/*
 *  * =====================================================================================
 * 
 *          Filename:  test2.c
 *   
 *          Description:  conventer from dec to 0101
 *     
 *          Version:  1.0
 *          Created:  14-05-11 10:03:25
 *          Revision:  none
 *          Compiler:  gcc
 *           
 *          Author:  Dr. Fritz Mehner (mn), [email protected]
 *          Company:  FH Südwestfalen, Iserlohn
 *             
 * =====================================================================================
  */

#include        <stdio.h>
#include        <stdlib.h>

int
main ( int argc, char *argv[] )
        {
            int nummer, uitkomst, teller;
            char uitkomst2[10];
            nummer = 2 ;
            uitkomst = 2 ;
            teller = 0 ;
            while (uitkomst > 0)
                {
                        uitkomst2[teller++]= uitkomst%2;
                        uitkomst = nummer /2 ;
                        nummer = uitkomst;
                        printf ("teller:%d  deling:%d uitkomst deling:%d  \n",teller, uitkomst%2, uitkomst);
                }
            printf ("teller is nu %d", teller);
            while (teller >  0)
                  {
                        printf("%d", uitkomst2[teller]);
                        teller = teller - 1;
                  }
  return EXIT_SUCCESS;
        }
/*c----------  end of function main  ---------- */

---

But i see something wierd.
The output is now this

Code:

---

teller:1  deling:1 uitkomst deling:1 
teller:2  deling:0 uitkomst deling:0 
teller is nu 241

---

So the loop is executed right but after that teller is a complete other number.

Roelof

That's because you forgot to print a new line after the number. teller is 2, then you print some uitkomst2 after it.

A bit more simplification...

Code:

---

            while (uitkomst > 0)
                {
                        uitkomst2[teller++]= uitkomst%2;
                        uitkomst = nummer /2 ;
                        nummer = uitkomst;
                        printf ("teller:%d  deling:%d uitkomst deling:%d  \n",teller, uitkomst%2, uitkomst);
                }

---

You can eliminate the intermediate variable uitkomst, which you are only using to flip values in nummer by operating directly on nummer, like this...

Code:

---

while (nummer > 0)
  { uitkomst2[teller++] = nummer %2;
    nummer /= 2 ; }

---

Again... you should always try to find the simplest solutions to a problem. I appreciate that this is all new to you, but the best time to form good habits is *before* you form the bad ones...

Hello,

I found this answer which is working :

Code:

---

/*
 *  * =====================================================================================
 * 
 *          Filename:  test2.c
 *   
 *          Description:  conventer from dec to 0101
 *     
 *          Version:  1.0
 *          Created:  14-05-11 10:03:25
 *          Revision:  none
 *          Compiler:  gcc
 *           
 *          Author:  Dr. Fritz Mehner (mn), [email protected]
 *          Company:  FH Südwestfalen, Iserlohn
 *             
 * =====================================================================================
  */

#include        <stdio.h>
#include        <stdlib.h>

int
main ( int argc, char *argv[] )
        {
            int nummer, uitkomst, teller;
            char uitkomst2[10];
            nummer = 100 ;
            uitkomst = 100 ;
            teller = 0 ;
            while (uitkomst > 0)
                {
                        uitkomst2[teller++]= uitkomst%2;
                        uitkomst = uitkomst /2 ;
                }
            printf ("Het getal %d is binair ", nummer);
            while (teller >  0)
                  {
                        printf("%d", uitkomst2[teller-1]);
                        teller = teller - 1;
                  }
  return EXIT_SUCCESS;
        }
/*c----------  end of funcrion main  ---------- */

---

Roelof

Edit 1 :

I saw later your solution.
I elimated number in the loops.
I use it only for displaying the answer.
But thanks for the help

CommonTater

Is my programm good or are there more things I have to learn,
If not, I will try to change this to converting to a octal number.

Roelof

Quote:

---

Originally Posted by roelof

CommonTater

Is my programm good or are there more things I have to learn,
If not, I will try to change this to converting to a octal number.

Roelof

---

Well... my thought would be there are always things we ALL have to learn. Programming is a never ending learning opportunity, a gift to those who love a challenge and the nemisis of those who don't.

Your code is working, which for a chapter 1 practice exercise is a pretty good achievement, especially in such a short time. You can be proud of it.

Improvement wise, not much left to work on... 100%....

So, on to the next chapter... don't stray too far from the ordered learning of your texts, they are organized in such a way as to lead you through the process in a well intentioned manner.

If you're thinking of changing it to print octal... Look at the formatters for printf()... I think you'll find you can reduce your code to about 5 lines. Seriously... they beat you to it... :D

Not too long ago there was a guy here who was trying to do decimal to hexidecimal conversion. This went on for several pages and lots of the guys were chipping in... Finally seeing that it was going noplace, and from my penchant for simplicity, I posted something very close to this...

Code:

---

#include <stdio.h>

int main (void)
  { unsigned int number;

    scanf("%d", &number);
    printf("\n0x%8X\n",number);

    return 0; }

---

The final conclusion was: "I'll bet that teacher never gives that assignment again." LOL...

Keep working my friend, you're doing just fine!

Hello,

I know that printf can do the job.
But because in chapter 1 this is not in the text I'm not allowed to use it.
But when looking at the prevoius exercises I have a idea how to do it.
And I think I have to adapt this source a little bit.

Roelof

Hello,

I adapted the script so it looks like this:

Code:

---

* 
 *          Filename:  test2.c
 *   
 *          Description:  conventer from dec to 0101
 *     
 *          Version:  1.0
 *          Created:  14-05-11 10:03:25
 *          Revision:  none
 *          Compiler:  gcc
 *           
 *          Author:  Dr. Fritz Mehner (mn), [email protected]
 *          Company:  FH Südwestfalen, Iserlohn
 *             
 * =====================================================================================
  */

#include        <stdio.h>
#include        <stdlib.h>

int
main ( int argc, char *argv[] )
        {
            int nummer, uitkomst, uitkomst3, uitkomst4, teller;
            char uitkomst2[10] ;
            nummer = 21 ;
            uitkomst = 21 ;
            teller = 0 ;
            while (uitkomst > 0)
                {
                        uitkomst3 = (uitkomst % 16) ;
                        if (uitkomst3 >=10)
                        {
                        uitkomst4 = uitkomst3 + 55 ;
                        }
                        else
                        {
                          uitkomst4 = uitkomst3 + 48 ;
                                }

                        uitkomst2[teller++]= uitkomst4;
                        uitkomst = uitkomst /16 ;
                }
            printf ("Het getal %d is octal ", nummer);
            while (teller >  0)
                  {
                        printf("%c", uitkomst2[teller-1]);
                        teller = teller - 1;
                  }
  return EXIT_SUCCESS;
        }

---

and Commentator thanks that you would be a friend.

Roelof

You might get a better mark, and you might solve minor bugs more easily, if you learn to indent code properly.