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Select the missing code lines from the dropdown provided.The C code given below should count digits of a number entered by the user.But, line numbers 7 and 9 are missing.1. #include <stdio.h>2. int main()3. {4.   int num,count=0;5.   printf("Enter an integer: ");6.   scanf("%d", &num);7.   ---------------Missing code -------------------8.   {9.   ---------------Missing code -------------------10.       ++count;11.   }12.   printf("Number of digits: %d",count);13. }

Question

Select the missing code lines from the dropdown provided.The C code given below should count digits of a number entered by the user.But, line numbers 7 and 9 are missing.1. #include <stdio.h>2. int main()3. {4.   int num,count=0;5.   printf("Enter an integer: ");6.   scanf("%d", &num);7.   ---------------Missing code -------------------8.   {9.   ---------------Missing code -------------------10.       ++count;11.   }12.   printf("Number of digits: %d",count);13. }

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Solution

The missing code lines are for a while loop that will continue as long as the number is not zero. In each iteration of the loop, the number is divided by 10, effectively removing the last digit. This is how the code counts the number of digits. Here are the missing lines:

  1. while(num != 0)
  2. num = num / 10;

So the complete code would be:

  1. #include <stdio.h>
  2. int main()
  3. {
  4. int num,count=0;
  5. printf("Enter an integer: ");
  6. scanf("%d", &num);
  7. while(num != 0)
  8. {
  9. num = num / 10;
  10.  ++count;
  11. }
  12. printf("Number of digits: %d",count);
  13. }

This problem has been solved

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Write a C program to count number of digits in a number. Using do-while loop

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