C Programming - Complicated Declarations - Discussion

i = (int)(char)(float) i;

Can any one explain how this typecasting works? Why there won't be any changes in its size? Its little confusing.

@Shivaraj.

Then also the output would have been 8.

The output will be different when you write the statement within sizeof as
printf("%d",sizeof((int)(float)(char)i));

What if after typecast the value of i is stored back is
i = (int)(char)(float)i;

Explanation:

#include<stdio.h>
double i;

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

If you compile above commented code it will not show any error because it took double size as 8 and display it.
Now,

#include<stdio.h>
//double i;

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

Now when we compile this code we will get an error.
Error: \'i\' undeclared (first use in this function),
Because the compiler ignoring this line.

I hope everyone can understand.

Thanks @ROHIT.

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

@Wikiok...Thanks for explanation

#include<stdio.h>
double i;

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

GUD EXPLANATION!.....@Kickem

(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)