Is there any difference between following declarations?

1 :	extern int fun();
2 :	int fun();

C data types classification are

1. Primary data types
   1. int
   2. char
   3. float
   4. double
   5. void
2. Secondary data types (or) User-defined data type
   1. Array
   2. Pointer
   3. Structure
   4. Union
   5. Enum

So, clearly long int l = 2.35; is not User-defined data type.
(i.e.long int l = 2.35; is the answer.)

A function prototype in C or C++ is a declaration of a function that omits the function body but does specify the function's name, argument types and return type.

While a function definition specifies what a function does, a function prototype can be thought of as specifying its interface.

- During declaration we tell the datatype of the Variable.

- During definition the value is initialized.

fmod(x,y) - Calculates x modulo y, the remainder of x/y.
This function is the same as the modulus operator. But fmod() performs floating point divisions.

Example:

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

int main ()
{
  printf ("fmod of 3.14/2.1 is %lf\n", fmod (3.14,2.1) );
  return 0;
}

Output:
fmod of 3.14/2.1 is 1.040000

Declaring is the way a programmer tells the compiler to expect a particular type, be it a variable, class/struct/union type, a function type (prototype) or a particular object instance. (ie. extern int i)

Declaration never reserves any space for the variable or instance in the program's memory; it simply a "hint" to the compiler that a use of the variable or instance is expected in the program. This hinting is technically called "forward reference".

/* Example for ceil() and floor() functions: */

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

int main()
{
    printf("\n Result : %f" , ceil(1.44) );
    printf("\n Result : %f" , ceil(1.66) );
 
    printf("\n Result : %f" , floor(1.44) );    
    printf("\n Result : %f" , floor(1.66) );

    return 0;
}
// Output:
// Result : 2.000000
// Result : 2.000000
// Result : 1.000000
// Result : 1.000000