printing number from array question..

i got this 50 cell integer array
i use only a few cells from 0(which represents the highest coefficient of 10^(col-1) )
to cell col-1 (which coefficient is 10^0 =1)
i got an input in this array
i could have the number 123 in cell col-1
so i need to add in cell col-1 the number 3
int cell col-2 (tens) i need to add 2
and in cell col-3 (the hundreds) i need to add 1

the problem is that in some cell there could be such a big number
that i would need to change the variable col.
and i couldnt do this simple process that i described in the beginning

i tried to build this :

Code:

---

                                     
kounter=0;
for (index=0;index<cols;index++){
    sumc[index]=0;
 }
for(index=cols-1;index>=cols;index--)
      {
                copy=sumc[index+kounter];
                  while (copy!=0){

                      sumc[index+kounter]=sumc[index+kounter]+copy%10;
                      copy=copy/10;
                      kounter--;
                    }
        }

 for (index = 0; index < rows; index++)
      {
          printf("%d",sumc[kndex]);

      }

---

so if i got this array

|0|0|0|8|19|10|

the array should look
|0|0|1|0|0|0|
it needs to print
1000

i cant calculate the sum because it could go over the integer bounders
it we are told to print the number as it should be and use integers only

??

Well you could store the number going from the ones place to the biggest place that you want to support, that would make calculations easier.

Code:

---

int input = 123;
int odometer[4] = {0}; /** largest number supported: 9999 **/
int i;
for( i = 0; i < 4 && input > 0; i++) {
  odometer[i] = input % 10;
  input /= 10;
}

---

And you would just reverse the array when you want to print.

when i entered this array
|0|0|0|8|19|10|

the array should look
|0|0|1|0|0|0|

i tried to fit your code
its not working

Code:

---

              for (kndex=0;kndex<cols;kndex++){
                  input=sumc[kndex];
            for( index = 0; index < cols && input > 0; index++) {
          sumc[index] = input % 10;
              input /= 10;
            }
              }

---

You have been told time and time again, and I am pretty sure that I have given you at least one example: indent your code properly!

Quote:

---

Originally Posted by transgalactic2

i tried to fit your code
its not working

---

How does it not work? The next time you merely say "its not working", I'm going to say "too bad" and laugh at your incompetence :p

If you want to avoid getting laughed at, read the article I linked to in my signature concerning how to ask smart questions.

Often, it would also help if you posted the smallest and simplest (compilable) program that demonstrates the problem. (Sounds familiar?)

this code is intended
and basicly what the code needs to do is
when i entered this array
|0|0|0|8|19|10|

the array should look
|0|0|1|0|0|0|

Code:

---

kounter=0;
for (index=0;index<cols;index++){
    sumc[index]=0;
 }
for(index=cols-1;index>=cols;index--)
      {
                copy=sumc[index+kounter];
                  while (copy!=0){

                      sumc[index+kounter]=sumc[index+kounter]+copy%10;
                      copy=copy/10;
                      kounter--;
                    }
        }

 for (index = 0; index < rows; index++)
      {
          printf("%d",sumc[kndex]);

      }

---

Quote:

---

Originally Posted by transgalactic2

this code is intended

---

Yes, but not properly (e.g., consistently). This code is indented properly:

Code:

---

kounter = 0;
for (index = 0; index < cols; index++)
{
    sumc[index] = 0;
}

for (index = cols - 1; index >= cols; index--)
{
    copy = sumc[index + kounter];

    while (copy != 0)
    {
        sumc[index + kounter] = sumc[index + kounter] + copy % 10;
        copy = copy / 10;
        kounter--;
    }
}

for (index = 0; index < rows; index++)
{
    printf("%d", sumc[kndex]);
}

---

Quote:

---

Originally Posted by transgalactic2

and basicly what the code needs to do is

---

Yes, you have explained that. Basically, you are trying to do some kind of bignum-style addition where each element of the array is a digit. What you have not explained is how does your current code not work.

It looks like you are missing a variable to carry over the addition.

Oh, and whiteflags suggested a little-endian format. Are you going to do it that way, or stick to a big-endian format?

i am going to try and fix my code
it work b taking a member
taking its remainder by 10
and adding this number to the first
then deviding it by 10
because its an int
it will keep only the whole part
and adding it remainder by 10 on the next cell
etc..

Quote:

---

Originally Posted by transgalactic2

it work b taking a member
taking its remainder by 10
and adding this number to the first
then deviding it by 10

---

That is not quite right. I would expect the algorithm to be something like this:

Code:

---

carry = 0
loop from the least significant "digit" to the most significant "digit"
    number = digit + carry
    digit = number % 10
    carry = number / 10 (integer division)

---

in what place i put values in the array?

Quote:

---

Originally Posted by transgalactic2

in what place i put values in the array?

---

What do you mean? Do you understand my pseudo-code?

i cant understand this scheme
its all about changing the array
and you didnt mention it in your "code"

Quote:

---

Originally Posted by transgalactic2

its all about changing the array
and you didnt mention it in your "code"

---

Can you see that each element of the array is a "digit"?

ok i will try to build a code uppon this scheme..

is that ok?

Code:

---

carry = 0
//loop from the least significant "digit" to the most significant "digit"
for(index=cols-1;index>=0;index++){
    number = suma[index] + carry;
    suma[index] = number % 10
    carry = number / 10
  }

---

Well... you have to make it syntactically correct C.

i tried to make it work
and it doesnt print anything
it just terminates the program
although i set the arra to be all zeros

??

Code:

---

                                          //start calculating the long addition procces

for (kndex = 0; kndex < cols; kndex++)
  {
      for (index = 0; index < rows; index++)
      {
         
        arr[kndex]=arr[kndex]+matrix[index][kndex];
          }

  }


carry = 0;
//loop from the least significant "digit" to the most significant "digit"
for(index=cols-1;index>=0;index++){
    number = arr[index] + carry;
    arr[index] = number % 10;
    carry = number / 10;
  }

for (kndex = 0; kndex < cols; kndex++)
      {
printf("%d ",arr[kndex]);
}

---

transgalactic2, indent your code properly. Post the smallest and simplest compilable code that demonstrates the problem.

I have been repeating this to you for a very, very long time. If you are not going to listen, I am not going to help.

this is the simplest code
the first part is taking the sum of all colums into one array
the second part is supposed to do the proccess i described in the start
i built the second part uppon your algorithm
and my main matrix was
1 1 1
2 2 2
so the array was
3 3 3
but when it enters your loop it should print 333
instead it going to endless loop
??

Code:

---

                                                            //start first part
for (kndex = 0; kndex < cols; kndex++)
  {
      for (index = 0; index < rows; index++)
        {
 
        arr[kndex]=arr[kndex]+matrix[index][kndex];
          }

    }          //end first part
                                                                                                                       
        carry = 0;          //start second part
      for(index=cols-1;index>=0;index++)
            {
            number = arr[index] + carry;
            arr[index] = number % 10;
            carry = number / 10;
            }

for (kndex = 0; kndex < cols; kndex++)
      {
            printf("%d ",arr[kndex]);
      }                                                      //end second part

---

Quote:

---

Originally Posted by transgalactic2

this is the simplest code

---

No, it is not the simplest code. If it is the simplest code then you are doing something wrong since the "first part" is completely wrong - there is no point computing the sum of all the columns into one array when you can create that one array from scratch. This is an example of what I am talking about:

Code:

---

#include <stdio.h>

int main()
{
    int digits[50] = {0};
    int i;
    int carry;
    digits[47] = 8;
    digits[48] = 19;
    digits[49] = 10;

    carry = 0;
    for (i = 49; i >= 0; --i)
    {
        int number = digits[i] + carry;
        digits[i] = number % 10;
        carry = number / 10;
    }

    for (i = 0; i < 50; ++i)
    {
        printf("%d", digits[i]);
    }
    putchar('\n');

    return 0;
}

---

Look at how I concentrate on the problem. Look at how you can compile and run the code as-is.

i need to shift the whole array the the right in order that proccess to work

my array looks like
|1|2|3|0|0|0|....|0|0|0|
i need it to look like |0|0|...|1|2|3|
i tried to do it like this
??

Code:

---

for (index = 0; index < 50; index++)
      {
        temp4=digits[49-(cols-1)+index];
        digits[49-(cols-1)+index]=digits[index];
                digits[index]=temp4;
      }

---

Quote:

---

Originally Posted by transgalactic2

i need to shift the whole array the the right in order that proccess to work

---

That does not make sense. The process should work as-is.

If you do want to "shift" it, how will that work? A right shift would mean discarding the least significant "digit".

your code is ok
it works fine
but my whole program works on
this |1|2|3|0|0|0|....|0|0|0|
kind of array for which i used only the |1|2|3| part
in order to do this operation like you showed in your code
i need it to look like |0|0|...|1|2|3|

Code:

---

/*      your array is shifted them to the right i need to do that to
  digits[47] = 8;
    digits[48] = 19;
    digits[49] = 10;
*/
for (index = 0; index < 50; index++)
      {
        temp4=digits[49-(cols-1)+index];
        digits[49-(cols-1)+index]=digits[index];
                digits[index]=temp4;
      }

---

Quote:

---

Originally Posted by transgalactic2

in order to do this operation like you showed in your code
i need it to look like |0|0|...|1|2|3|

---

Ah, but you don't. In your case you are using a little-endian format, so instead of looping in reverse, loop in a forward direction. The algorithm is otherwise the same.

the problem is with the zeros
if my number is
3|3|3|0|000.........
i need to cut the zeros
so i could say if you see a zero then stop printing
but what if
my array is
3|0|3|0|000......... in that case it whould print only 3 instead of 303
thats why i need it to shift the the right like you did
the problem with that i am afraid that if i will get all the cells full
then by "shifting" it will change the number i am suppossed to print
??

Code:

---

for (index = 0; index < 50; index++)
      {
       
        digits[49-(cols-1)+index]=digits[index];
               
      }

---

when i copy the value to the other end i need to put other value to cell i just copied from
if i put zero it will work for such numbers as |3|3|3|....|0| =>|0|0|0|....|3||3||3|
but if i get a number like all 9
|9|9|9|...|9| then it would change the number instead of shifting..
what to do with the cell i just copied from
???

Quote:

---

Originally Posted by transgalactic2

so i could say if you see a zero then stop printing

---

When printing, scan from the right until you find a non-zero, then keep printing until the end. Basically, you are just ignoring insignificant zeroes. If you scan to the end (i.e., the first element, all the way on the left) without finding a non-zero, the number itself is zero.

when i tried to do your example on my type of array
and i got a line of zeros
how to shift it to the left if we get a number that goes out of the
boundres
because our number was 1000
and we had only 3 places
??

Code:

---

digits[0] = 8;
    digits[1] = 19;
    digits[2] = 10;

         
carry = 0;
    for (index = 49; index >= 0; --index)
    {
        int number = digits[index] + carry;
        digits[index] = number % 10;
        carry = number / 10;
    }

    for (index = 0; index < 50; ++index)
    {
        printf("%d", digits[index]);
    }
    putchar('\n');

---

Quote:

---

Originally Posted by transgalactic2

when i tried to do your example on my type of array
and i got a line of zeros

---

I notice that you are still looping in reverse.

Quote:

---

Originally Posted by transgalactic2

how to shift it to the left if we get a number that goes out of the
boundres
because our number was 1000
and we had only 3 places

---

As I said, do not attempt to "shift". Just ignore the leading zeroes, e.g.,

Code:

---

#include <stdio.h>

#define NUM_DIGITS 50
int main()
{
    /* Use little-endian format. */
    int digits[NUM_DIGITS] = {10, 19, 8};
    int i;
    int carry;

    /* Normalise such that each digit is a numeral in base 10. */
    carry = 0;
    for (i = 0; i < NUM_DIGITS; ++i)
    {
        int number = digits[i] + carry;
        digits[i] = number % 10;
        carry = number / 10;
    }

    /* Find the first non-zero. */
    for (i = NUM_DIGITS - 1; i > 0 && digits[i] == 0; --i);

    /* Print the significant digits. */
    for (; i >= 0; --i)
    {
        printf("%d", digits[i]);
    }
    putchar('\n');

    return 0;
}

---

i get 000000000000000000000000000000000000....00001000
when i run your code
i need only 1000
??

Then you broke laserlight's code. If you run it, exactly as written, you will actually get 1000.

Quote:

---

Originally Posted by transgalactic2

i get 000000000000000000000000000000000000....00001000
when i run your code
i need only 1000
??

---

I get 1000 when I run the program whose source code I showed. If you get the leading zeroes, it implies that you are not running my most recent example.

More importantly, you need to understand what the code commented by "Find the first non-zero." and "Print the significant digits." does.

i fixed it

why when i put

Code:

---

int digits[50]={9, 1, 100};
it gives me
10019
??

---

i changed your code to

Code:

---

/* Find the first non-zero. */
    for (i = NUM_DIGITS - 1; i > 0 && digits[i] == 0; --i);
 temp4=i;
    /* Print the significant digits. */
    for (i=49; i >= 0; --i)
    {
                if (temp4>=i){
        printf("%d", digits[i]);
                }
    }
    putchar('\n');

---

Quote:

---

Originally Posted by transgalactic2

why when i put

Code:

---

int digits[50]={9, 1, 100};
it gives me
10019
??

---

---

What output did you expect?

Quote:

---

Originally Posted by transgalactic2

i changed your code to

---

What do you hope to achieve by making such a change?

i expect
900 +10 +100
1010

??

Since the first 9 means 9 and not 900 that seems like a rather silly expectation.

Quote:

---

Originally Posted by transgalactic2

i expect
900 +10 +100
1010

---

This is frustrating. It looks like you cannot decide how you really want to interpret the array.

At this point I would like to ask what exactly are you trying to do. What is the point of this whole exercise?

10^2 10^1 10^0 etc..

Quote:

---

Originally Posted by transgalactic2

10^2 10^1 10^0 etc..

---

That is too vague.

Quote:

---

Originally Posted by transgalactic2

10^2 10^1 10^0 etc..

---

Yes. And since you start at the right......................