having a problem with types

I'm looking for the operator input to be differentiated from integer and float. Here is what I have so far (which doesn't seem to work)

#include<stdio.h>
#include<stdlib.h>

int num1, num2, num3;
int main (void)
{
printf("Input a number:");
scanf("%d",&num1);

num2=(num1)/2; /* outputs an integer number if float is entered */
num3=num2*2; /* should not equal num1 if float number was entered */

if (num3==num1)
printf("\nThis is an integer");
else
printf("\nThis is a float");


return 0;
}

Please someone tell me where I'm going wrong...I appreciate the help.

>int num1, num2, num3;

You need to declare num1 and num3 as float or double:
float num1, num3;

I tried that, but I get the answer "float" all the time. It won't discern the integer numbers.

The best this code will tell you is whether a number is even or odd.

> scanf("%d",&num1);
%d will stop scanning at a decimal point, so an input of
100.3
or
100
will both result in num1 being 100

> I'm looking for the operator input to be differentiated from integer and float.
Then you need to read the input as a string, and check that
- it contains only digits and perhaps a decimal point
- if there is a '.', then its a float, otherwise its an int

You could also read it in always as a float, and check to see if the fractional part is zero.

E.g.:

double number;
scanf("%lg",&number);
if (fmodl(number,1.0) != 0) printf("Float!\n");
else printf("Int!\n");

BTW, fmodl is not ANSI C. fmod, however, is.

So this will work, using single-precision floats:

float number;
scanf("%g",&number);
if (fmod(number,1.0) != 0) printf("Float!\n");
else printf("Int!\n");

This works a little better than simply scanning for deciamls, because:

2.00, 1.02e4, etc. are actually ints (represented as floating point values). This code will allow you to know if the actual value is an integer or not. If you want to know if the representation is a floating point one (i.e. you want "2" to return int, but "2.0" to return float) you'll need to scan for the decimal point as Salem said.

Try this:

Code:

---

int main (void)
{
float num1, num2;
printf("Input a number:");
scanf("%f",&num1);
num2=(int)(num1*10.)/10;
if (num2==num1)
  printf("\nThis is an integer");
else
  printf("\nThis is a float");
return 0;
}

---

Hey! It works! I understand the use of the cast in the math expression, but what I don't understand is why would you multiply by "10."? (i.e. (int)(num1*10.)/10)

I'll try to explain.
Any real number (float) is always written in base 10. This is the reason for multiplying by 10, instead of 2. We don't really care how the real number is represented internally. So for example:

5.3 * 10 = 53
53 / 10 = 5.3
(int) 5.3 = 5

5 * 10 = 50
50 / 10 = 5
(int) 5 = 5

This is similar to the nonfunction method of finding the remainder(modulus):
Using integer arithmetic:
remainder = number - (number/divisor*divisor)
remainder = 34 - (34/10*10) = 34 - 30 = 4
remainder = 34 - (34/2*2) = 0

Your code looked very similar to this.

The fmodl() solution posted by The V. is a good way to do it. I generally use library functions when I can.