C Programming - Declarations and Initializations - Discussion

extern int fun(float); this is the declaration.

int fun(int aa)
This is the function defined.

But both the prototype is not matching.
fun(float) fun(int) so error.

If fun(int) is replaced with fun(float)then 3.14 is assigned to aa and returns 3 after that ++a.

So 3+1=4.

I didn't understand that how 3.14 becomes 3?

And then we are adding 1 to it.

Becz according to me we should add 3.14+1=4.14.

But the correct answer is 4.

Please clear my doubt how 3.14 becomes 3 first?

@Neha.

After increment we get 4.14.

But when it return it will be type casted to int so 4 will get stored in a.

There is no need of explicit typecasting because the return type is int.

#include<stdio.h>
int main()
{
extern int fun(float);
/*here we inform to the compiler that we are sending a float */argument
int a;
a = fun(3.14);
/*here we are successfully sent the float value as argument
*/
printf("%d\n", a);
return 0;
}
int fun(int aa)
/*but here it is defined as, it will receive integer argument
we are violating the declaration so it is shows that errors */
{
return (int)++aa;
}