simple 5 digit

**wonderpoop** · 10-15-2006

I need to have a user type in 5 single digit numbers spaced, which works, but how can i fix this so if they type in more then 5 digits, or try to type in a 2 digit number. For example 12345 should look like 1 2 3 4 5.

I want to fix this so that if they type in 1112131415 it will give a message.

Code:

 float choice;

  printf("Enter a five-digit number:");
  scanf("%d", &num);

  if ((choice == 0) || (choice == 9))
    {
  num1 = ((num % 10) / 1);
  num2 = ((num % 100) / 10);
  num3 = ((num % 1000) / 100);
  num4 = ((num % 10000) / 1000);
  num5 = ((num % 100000) / 10000);
  printf("%d  %d  %d  %d  %d\n", num5, num4, num3, num2, num1);
    }
  else
    {
      printf("Enter five single digit numbers 0 thru 9\n");
    }
  return 0;
}

**Happy_Reaper** · 10-15-2006

If (num/100000 > 1) you have a 6 digit number. If (num/10000) < 1, you have less then 5 digits.

**yxunomei** · 10-15-2006

Originally Posted by Happy_Reaper

If (num/100000 > 1) you have a 6 digit number. If (num/10000) < 1, you have less then 5 digits.

shouldn't it be

Code:

if(num/100000 >= 1)

with your original code, if I type a number between 100000 to 199999, it will be considered as 5 digits...

**wonderpoop** · 10-15-2006

Originally Posted by yxunomei

shouldn't it be

Code:

if(num/100000 >= 1)

with your original code, if I type a number between 100000 to 199999, it will be considered as 5 digits...

Thanks

**Thantos** · 10-15-2006

Code:

#include <stdio.h>

int main()
{
  unsigned test_inputs[] = { 12345, 12, 1234567, 123, 1 };
  int i = 0, num_inputs = 5;
  char buffer[BUFSIZ];

  for(i = 0; i < num_inputs; i++)
  {
    sprintf(buffer, "%u", test_inputs[i]);
    printf("%u %s a 5 digit number\n", test_inputs[i], strlen(buffer) == 5 ? "is" : "is not");
  }
  return 0;
}

**Happy_Reaper** · 10-15-2006

Originally Posted by yxunomei

shouldn't it be

Code:

if(num/100000 >= 1)

with your original code, if I type a number between 100000 to 199999, it will be considered as 5 digits...

Right...I guess we're even.

**wonderpoop** · 10-15-2006

I did it this way, something we are learning as well, this works, but again, now I think I need to change my if statement. It works, but it is still buggy for example:

[academ] $ digit5.out
Enter a five-digit number:12345
1 2 3 4 5

[academ] $ digit5.out
Enter a five-digit number:1112131415
Enter four single digit number

[academ] $ digit5.out
Enter a five-digit number:123456
2 3 4 5 6

Code:

#include <stdio.h>
#include <math.h>

int main()
{
  int num, num1, num2, num3, num4, num5;
  int rm1, rm2, rm3, rm4, rm5;
 
  printf("Enter a five-digit number:");
  scanf("%d", &num);

  if (num/10000 <= '1')
    {
  num1 = ((num % 10) / 1);
  rm1 = ((num1++) % 10);
  num2 = ((num % 100) / 10);
  rm2 = ((num2++) % 10);
  num3 = ((num % 1000) / 100);
  rm3 = ((num3++) % 10);
  num4 = ((num % 10000) / 1000);
  rm4 = ((num4++) % 10);
  num5 = ((num % 100000) / 10000);
  rm5 = ((num5++) % 10);
  printf("%d  %d  %d  %d  %d\n", rm5, rm4, rm3, rm2, rm1);
    }
  else
    {
  printf("Enter five single digit numbers 0 thru 9\n");
    }
  return 0;
}

**wonderpoop** · 10-15-2006

I now did it this way, but the if isn't completely working. If i put in 12345 I get 1 2 3 4 5, if i put in 1112131415 I get the else statement, but if i put in 123456 I get 2 3 4 5 6.

Code:

#include <stdio.h>
#include <math.h>

int main()
{
  int num, num1, num2, num3, num4, num5;
  int rm1, rm2, rm3, rm4, rm5;
 
  printf("Enter a five-digit number:");
  scanf("%d", &num);

  if (num/10000 <= '1')
    {
  num1 = ((num % 10) / 1);
  rm1 = ((num1++) % 10);
  num2 = ((num % 100) / 10);
  rm2 = ((num2++) % 10);
  num3 = ((num % 1000) / 100);
  rm3 = ((num3++) % 10);
  num4 = ((num % 10000) / 1000);
  rm4 = ((num4++) % 10);
  num5 = ((num % 100000) / 10000);
  rm5 = ((num5++) % 10);
  printf("%d  %d  %d  %d  %d\n", rm5, rm4, rm3, rm2, rm1);
    }
  else
    {
  printf("Enter five single digit numbers 0 thru 9\n");
    }
  return 0;
}

**peterchen** · 10-15-2006

check if the number is between 9999
and 100000 exclusive

Code:

if(input > 9999 && input < 100000) {
 /* your code continues */

a better way is to store the input as a string instead of an int. that way overfloat can be detected. basically all u have to do is check that all the contents of the array are digits and that the string is of length 5. then index through the array to print out the digits

see code

Code:

#include <stdio.h>
#include <string.h>

int main(void) {
    char nums[7];
    int i;
    int flag=0;
    
    printf("Enter five digit number: ");
    scanf("%s", nums);
    
    /* check contents of  array */
    if(strlen(nums)==5) flag=1;
    for(i=0; i<5 && flag; i++) {
        if(nums[i] <='0' || nums[i] >= '9') flag =0;
    }
    
    /* print response */
    if(flag) {
        for(i=0; i<5; i++) printf("%c ", nums[i]);
    } else {
        printf("Input did not contain five digits\n");
    }
    system("PAUSE");
    return 0;
}

**yxunomei** · 10-16-2006

Originally Posted by Happy_Reaper

Right...I guess we're even.

LOL ...

... your funny one

**yxunomei** · 10-16-2006

Code:

#include <stdio.h>
#include <math.h>

int main()
{
  int num, num1, num2, num3, num4, num5;
  int rm1, rm2, rm3, rm4, rm5;
 
  printf("Enter a five-digit number:");
  scanf("%d", &num);

  if (num/10000 <= '1') //  wrong!  see http://www.lookuptables.com/
                        //  you are doing if(num/10000 <= 49) instead of 1 
    {
  num1 = ((num % 10) / 1);
  rm1 = ((num1++) % 10);
  num2 = ((num % 100) / 10);
  rm2 = ((num2++) % 10);
  num3 = ((num % 1000) / 100);
  rm3 = ((num3++) % 10);
  num4 = ((num % 10000) / 1000);
  rm4 = ((num4++) % 10);
  num5 = ((num % 100000) / 10000);
  rm5 = ((num5++) % 10);
  printf("%d  %d  %d  %d  %d\n", rm5, rm4, rm3, rm2, rm1);
    }
  else
    {
  printf("Enter five single digit numbers 0 thru 9\n");
    }
  return 0;
}

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