C Programming - Complicated Declarations - Discussion

Can any one explain this?

In the line <(int)(float)(char) i;> the casts do not modify the type of the variable i, so i is still a Double stored 8 bytes.

Initially i is declared as double. So i=8 .

What is the difference between these two
(float) i; and
float i;

It produce the size as 8 and 4 respectively.?

(float) i : Is not a declaration, but a cast. You need an lvalue too. It doesn't change the sizeof(i).
float i : It is a declaration. sizeof(i) = 4 (32bit platform)

GUD EXPLANATION!.....@Kickem

#include<stdio.h>
double i;

int main()
{
(int)(float)(char) i;
printf("%d",sizeof(i));
return 0;
}

@Wikiok...Thanks for explanation

I am satisfied with Kickem's explanation.

In this program the line
(int)(float)(char) i;........... is useless

If you compile the program then find that compiler say's that "Code has no effect" and highlight this line.

In this program the compiler print the size of double which is declared global in the program.

If you want to check that how it is true then you will replace double to the int or float you find that it print's the size of int or float.

Write the program like

#include<stdio.h>
int i;

int main()
{
(int)(float)(char) i;
printf("%d",sizeof(i));
return 0;
}

Output is : 2

Thank you friends if you have more questions the contact me in gmail.com my email-id is : rk52567(at)gmail.com

Thanks @ROHIT.