Question

I need to write a C program that will compare the number of digits before decimal point and after the decimal point and make sure they are equal.

How can I count how many powers of ten we have before and after the decimal point?

Here is what I have so far:

void main()
{
    is_equal(6757.658);   
}

INT is_equal(double x)
{
   int digits = 0;
   while (x) {
     x /= 10;
     digits++;
   }

   printf("%d ",digits);
}

Is there a better way to do this?

Answer 1

3rd try:

Count the number of “digits” before and after a “.”

• Null is considered not equal to anything

• I did not test this code it might contain typos.

  int is_equal(char *buffer)
  {
    char *temp;
    int leftLen,rightLen;
  
    temp = strtok(buffer,".");
    if (temp == null) return false;
  
    leftLen = strlen(temp);
  
    temp = strtok(buffer,".");
  
    if (temp == null) return false;
  
    rightLen = strlen(temp);
  
    return (leftLen == rightLen);
  }
  

---

Old stuff…

There are going to be lots of problems here, a floating point (double) in C is not always accurate to 100%; If you perform multiplication or division. If you multiply the digits will change.

The best way to solve this problem is to render the double to a string and then parse that string.

You can use sprintf to write the formatted double to a buffer.

OR

You can skip using a double all together and use a string to start with.

Thus building on Marcelo’s answer:

Read the string from the user into a buffer called buff

Then parse it with a statement like sscanf(buf, "%d.%d", &a, &b);

buff is a char * or a char [], a and b are int. You test by saying a == b

void main()
{
  is_equal("6757.658");   
}

int is_equal(char *x)
{
   int left,right;

   sscanf(x, "%d.%d", &left, &right);

   printf("Left digits: %d\n\r",left);
   printf("Right digits: %d\n\r",right);

   return (left == right);
}