adding a number to a number

This is a discussion on adding a number to a number within the C Programming forums, part of the General Programming Boards category; this is exactly what im trying to do, and its probably an easy question to solve. Say i have a ...

  1. #1
    Registered User
    Join Date
    Oct 2007
    Posts
    118

    adding a number to a number

    this is exactly what im trying to do, and its probably an easy question to solve. Say i have a number in a variable (ie. 1) and i want to add a 0 to the one (ie. 10) how would i go by doing this?

    I have to make a decimal to binary converter, so id need to now how to add a number to a number variable.

  2. #2
    Captain Crash brewbuck's Avatar
    Join Date
    Mar 2007
    Location
    Portland, OR
    Posts
    7,246
    Quote Originally Posted by bigmac(rexdale) View Post
    this is exactly what im trying to do, and its probably an easy question to solve. Say i have a number in a variable (ie. 1) and i want to add a 0 to the one (ie. 10) how would i go by doing this?

    I have to make a decimal to binary converter, so id need to now how to add a number to a number variable.
    Try to be more specific in your terminology. "Adding numbers" is accomplished with the addition operator "+". What you are trying to do is append a digit.

    Your approach seems to be to hold the binary "value" as an integer whose decimal representation consists only of 0 and 1. That is definitely not going to work, because integer values have limited range and you are wasting most of it by working in the wrong base.

    Numbers never have to be converted to/from specific bases while internally represented. A value is a value, not a sequence of digits. The digits come into being when the value is represented in generated output.

  3. #3
    Registered User
    Join Date
    Oct 2007
    Posts
    118
    OK, well this is what I've been thinking of doing.

    The program is Integer to Binary converter

    I use the integer number to be converted, in a function that does something like this:

    for loop(starting at 32, and descending, called counter)
    number = number - 2^counter(number is the number the user wants to be outputted to look like the binary equivalent, counter would be 32, then 31 ...)

    a if statement, where if number is less than 0, i add back what i substracted and the first digit of the binary output would be 0 (the variable name is b)

    else if, number is more than 0 or eqal to, the first digit of the binary output would be 1(the variable name is b)

    binary(a variable) would equal b

    is if the first line is true, it would equal 0, if not, then 1, but:

    then it does a next loop with counter being 31, so, how would the next digit (either a 0 or a 1) be appended? to what the digit was before.

    ie. if it was 1 when counter equaled 32, when it equals 31 and the answer is one again, i need the binary variable to equal '11'

  4. #4
    Registered User
    Join Date
    Nov 2006
    Posts
    42
    I'm not sure if I understand you correctly, but if you want to convert int to displayable binary, one quick approach would be to store remaining parts of division (which is either 1 or 0) in char array, eg.

    Code:
    int i = 0;
    while ( num > 0 ) { // num == int that'd be converted to binary
      if ( num % 2 == 0 ) 
          charArray[i] = '0';
      else 
          charArray[i] = '1';
      num /= 2;
      i++;
    }
    ...and then simply reverse array and print output.

  5. #5
    Registered User
    Join Date
    Oct 2007
    Posts
    118
    i didn't learn that yet though, should've mention im a beginner

    basically, im looking for a way to combine two numbers, i want to combine a int variable that is 2, with a next variable that is 5 to make the variable now equal 25

  6. #6
    Ex scientia vera
    Join Date
    Sep 2007
    Posts
    478
    Quote Originally Posted by bigmac(rexdale) View Post
    i didn't learn that yet though, should've mention im a beginner

    basically, im looking for a way to combine two numbers, i want to combine a int variable that is 2, with a next variable that is 5 to make the variable now equal 25
    And what exactly does that have to do with the binary numeral system?

    Using some simple arithmetic can accomplish what you want.

    25 = 2 * 10 + 5 * 1.

    So, basically..

    int a = 2, b = 5;

    a *= 10;
    a += b*1;

    Multiplying by one is pointless, but it's to clarify the concept.

  7. #7
    Registered User
    Join Date
    Oct 2007
    Posts
    118
    ok, so i changed it up, since it is a pointless code

    Code:
    #include<stdio.h>
    int binar(int number, int *pbinary)
    {
        *pbinary = number &#37; 2;
        return number / 2;
    }
    
    main () {
         int number, binary, bin;
         printf("number ");
         scanf("%d", &number);
         binary = binar(number, &bin);
         printf("%d %d", binary, bin);
    }
    i that that so far....I think i need to add an array and a for loop to proceed to get the rest of the binary digits

    edit:

    Code:
    #include<stdio.h>
    int binar(int number, int *pbinary)
    {
        *pbinary = number % 2;
        return number / 2;
    }
    
    main () {
         int number, binary, bin, bina[32], counter;
         printf("number ");
         scanf("%d", &number);
         while (number > 0 || number < 2147483647) {
         counter = counter + 1;
         binary = binar(number, &bin);
         bina[counter] = bin;
         }
         printf("%d %d", bina[1]);
    }
    i now have that, but its not displaying any text
    Last edited by bigmac(rexdale); 10-23-2007 at 02:19 PM.

  8. #8
    Registered User
    Join Date
    Nov 2006
    Posts
    42
    Okay, maybe I could help here:

    Code:
    int main (void)  // see FAQ entry here
    {
         int number, bin, counter = 0;  // you're starting accessing array at first index, which is 0
         char bina[32];  // storing array was supposed to be char type, so it can be displayed as a string
         printf("number ");
         scanf("%d", &number);
         while (number > 0 || number < 2147483647) // second condition isn't really needed
         { 
            number = binar(number, &bin);  // you need to modify original number that's checked in while loop
            bina[counter] = bin + '0';  // you don't want to put 1 or 0 into array, but it's character equivalent in ascii code
            counter = counter + 1; // increment counter once you're done with it's current value, not before
         }
         bina[counter] = 0;  // c-string should end with 0
         printf("%d %s", number, bina);
    }
    Don't forget to reverse array before final displaying (because what you got now will display from bits from end - to - begining) - also watch out for terminating 0 when reversing, you don't want it at the very begining of your string.
    Hope it's more clear now.

  9. #9
    Registered User
    Join Date
    Oct 2007
    Posts
    118
    i fixed it up and reversed the arrays (i have a differen code from the last time though)

    Code:
    #include<stdio.h>
    int binar(int number, int *pbinary)
    {
        *pbinary = number / 2;
        return number &#37; 2;
    }           
    
    main () {
         int binary, bin, bina[32], counter;
         int number, number2, numcheck;
         printf("number ");
         scanf("%d", &number);
         
         while (number < 0 || number > 2147483647) {
               printf("Enter Valid Number\n");
               printf("number ");
               scanf("%d", &number);
               }
         number2 = number;
         binary = binar(number, &bin);
         bina[1] = binary; 
         number = bin;
         bina[2] = binar(number, &bin);
         number = bin;
         bina[3] = binar(number, &bin);
         number = bin;
         bina[4] = binar(number, &bin);
         number = bin;
         bina[5] = binar(number, &bin);
         number = bin;
         bina[6] = binar(number, &bin);
         number = bin;
         bina[7] = binar(number, &bin); 
         number = bin;
         bina[8] = binar(number, &bin);
         number = bin;
         bina[9] = binar(number, &bin);
         number = bin;
         bina[10] = binar(number, &bin);
         number = bin;
         bina[11] = binar(number, &bin);
         number = bin;
         bina[12] = binar(number, &bin);
         number = bin;
         bina[13] = binar(number, &bin);
         number = bin;
         bina[14] = binar(number, &bin);
         number = bin;
         bina[15] = binar(number, &bin);
         number = bin;
         bina[16] = binar(number, &bin);
         number = bin;
         bina[17] = binar(number, &bin);
         number = bin;
         bina[18] = binar(number, &bin); 
         number = bin;
         bina[19] = binar(number, &bin);
         number = bin;
         bina[20] = binar(number, &bin);
         number = bin;
         bina[21] = binar(number, &bin);
         number = bin;
         bina[22] = binar(number, &bin);
         number = bin;
         bina[23] = binar(number, &bin);
         number = bin;
         bina[24] = binar(number, &bin); 
         number = bin;
         bina[25] = binar(number, &bin);
         number = bin;
         bina[26] = binar(number, &bin);
         number = bin;
         bina[27] = binar(number, &bin);
         number = bin;
         bina[28] = binar(number, &bin);
         number = bin;
         bina[29] = binar(number, &bin);
         number = bin;
         bina[30] = binar(number, &bin);
         number = bin;
         bina[31] = binar(number, &bin);
         number = bin;
         bina[32] = binar(number, &bin);
         printf("Integer:         32-bit binary:\n");
         printf("----------       --------------\n");
         printf("%10d%8d%d%d%d%d%d%d%d %d%d%d%d%d%d%d%d %d%d%d%d%d%d%d%d %d%d%d%d%d%d%d%d", 
    number2, bina[32], bina[31], bina[30], bina[29], bina[28], bina[27], bina[26], bina[25], bina[24], bina[23], 
    bina[22], bina[21], bina[20], bina[19], bina[18], bina[17], bina[16], bina[15], bina[14], bina[13], 
    bina[12], bina[11], bina[10], bina[9], bina[8], bina[7], bina[6], bina[5], bina[4], bina[3], bina[2], bina[1]);
    
    }
    edit: no problems i fixed it

    and a reply aobut the line of code:
    Code:
    while (number > 0 || number < 2147483647)
    the number has to be between 0 to 2147483647

    I just have a question now, i wrote so many lines of the same code (the arrays), would a for loop work so it wouldn't be so many lines of the same code
    Last edited by bigmac(rexdale); 10-24-2007 at 08:26 AM.

  10. #10
    Frequently Quite Prolix dwks's Avatar
    Join Date
    Apr 2005
    Location
    Canada
    Posts
    8,046
    the number has to be between 0 to 2147483647
    Yes, but an int can't hold a number larger than 2147483647. So checking for it is redundant. It's like saying, if the number is less than or equal to its largest possible number. It will always be true.

    I just have a question now, i wrote so many lines of the same code (the arrays), would a for loop work so it wouldn't be so many lines of the same code
    Yes, definitely. I was going to suggest it myself. Use a for loop with something like
    Code:
         binary = binar(number, &bin);
         bina[x] = binary;
    in the body, with x looping from 0 to 31, inclusive (x=0, x<32). You have from 1 to 32 inclusive, but that's causing a buffer overrun.

    [edit] I'd use a for loop for that last printf as well. [/edit]
    dwk

    Seek and ye shall find. quaere et invenies.

    "Simplicity does not precede complexity, but follows it." -- Alan Perlis
    "Testing can only prove the presence of bugs, not their absence." -- Edsger Dijkstra
    "The only real mistake is the one from which we learn nothing." -- John Powell


    Other boards: DaniWeb, TPS
    Unofficial Wiki FAQ: cpwiki.sf.net

    My website: http://dwks.theprogrammingsite.com/
    Projects: codeform, xuni, atlantis, nort, etc.

  11. #11
    Registered User
    Join Date
    Oct 2007
    Posts
    118
    ok so,

    Code:
    for (x = 0; x<32;) {
         binary = binar(number, &bin);
         bina[x] = binary;
    }
    would that do it?

    and for the for loop for the last printf, i have to make it also line up with the printf above it, so how would that be done?

    edit: i got it now, but for the bin[32] array it gives me a big number that i dont know what it stands for, so i multiplied it by 0 since it will always equal 0

    Code:
    #include<stdio.h>
    int binar(long number, int *pbinary)
    {
        *pbinary = number / 2;
        return number &#37; 2;
    }           
    
    main () {
         int binary, bin, bina[32], x;
         long number, number2;
         printf("number ");
         scanf("%d", &number);
         
         while (number < 0 || number > 2147483647) {
               printf("Enter Valid Number\n");
               printf("number ");
               scanf("%d", &number);
               }
         number2 = number;
         binary = binar(number, &bin);
         bina[1] = binary; 
         number = bin;
         for (x = 2; x < 32; x = x + 1) {
         bina[x] = binar(number, &bin);
         number = bin;
         }
         printf("Integer:         32-bit binary:\n");
         printf("----------       --------------\n");
         printf("%10d%8d%d%d%d%d%d%d%d %d%d%d%d%d%d%d%d %d%d%d%d%d%d%d%d %d%d%d%d%d%d%d%d"
    , number2, bina[32]*0, bina[31], bina[30], bina[29], bina[28], bina[27], bina[26], bina[25], bina[24], bina[23], 
    bina[22], bina[21], bina[20], bina[19], bina[18], bina[17], bina[16], bina[15], bina[14], bina[13], bina[12],
     bina[11], bina[10], bina[9], bina[8], bina[7], bina[6], bina[5], bina[4], bina[3], bina[2], bina[1]);
    Last edited by bigmac(rexdale); 10-24-2007 at 12:52 PM.

  12. #12
    Frequently Quite Prolix dwks's Avatar
    Join Date
    Apr 2005
    Location
    Canada
    Posts
    8,046
    Code:
    x = x + 1
    is the same as
    Code:
    x += 1
    or
    Code:
    x ++
    It's common practise to use x++. You don't have to, of course, but it's shorter and arguably more readable.

    edit: i got it now, but for the bin[32] array it gives me a big number that i dont know what it stands for, so i multiplied it by 0 since it will always equal 0
    That's because bina[32] doesn't exist.

    When you declare an array like this:
    Code:
    type array[N];
    you get N elements. Except they start counting from zero. So if N=3, you get 0, 1, and 2 as elements. That's still 3 elements, but the last element is N-1. So if you declare an array as
    Code:
    int bina[32];
    then your elements are 0, 1, 2, ... 31. Not 1, 2, 3, ... 32.

    I'm sure you could work out how to convert the last printf() into a loop. It would be a version of this:
    Code:
    for(...) {
        printf("%d", num[x]);
    }
    dwk

    Seek and ye shall find. quaere et invenies.

    "Simplicity does not precede complexity, but follows it." -- Alan Perlis
    "Testing can only prove the presence of bugs, not their absence." -- Edsger Dijkstra
    "The only real mistake is the one from which we learn nothing." -- John Powell


    Other boards: DaniWeb, TPS
    Unofficial Wiki FAQ: cpwiki.sf.net

    My website: http://dwks.theprogrammingsite.com/
    Projects: codeform, xuni, atlantis, nort, etc.

Popular pages Recent additions subscribe to a feed

Similar Threads

  1. Adding a Large number digit by digit
    By mejv3 in forum C Programming
    Replies: 23
    Last Post: 09-21-2007, 03:00 PM
  2. Adding a Large number digit by digit
    By mejv3 in forum C Programming
    Replies: 1
    Last Post: 09-14-2007, 03:28 AM
  3. Prime number program problem
    By Guti14 in forum C Programming
    Replies: 11
    Last Post: 08-06-2004, 04:25 AM
  4. help with a source code..
    By venom424 in forum C++ Programming
    Replies: 8
    Last Post: 05-21-2004, 12:42 PM
  5. parsing a number
    By juancardenas in forum C Programming
    Replies: 1
    Last Post: 02-19-2003, 12:10 PM

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21