C Programming - Declarations and Initializations - Discussion

Why including only 512?

Kindly help by explaining this.

When we are changing char array to short array it's not doing the same thing. In my compiler size of int is 4 and short is 2.

Please check by running the same program with c compiler, the Garbage value will be printed for i.

How/why we get 512 and 515? Please explain me.

But why are assuming that "i" will have size of union and not garbage value?

Because as I have studied about union, the most recent allocation is given the memory and rest all have garbage.

Every time I run the program, it didn't give 515 for me. It's giving some garbage value. Please help me.

Why u.i would print size of the union? What exactly u.i means?

How can we get the size of the entire union by u.i?

union a u;
printf("%d, %d, %d\n", u.ch[0], u.ch[1], u.i);

Can anyone explain these both lines?

Explanation to the given question is as follows. As we know union size is the biggest size of its element, hence integer is considered to be the biggest.

Here integer is considered to be 4 bytes which are equal to 32 bits (GCC compiler). The memory allocation is as follows. 0000000000000000000000010000000011.

Hence from the above figure u.ch[0] = 3 & u.ch[1]=2.

As the character is of 1 byte.

So in that 4 bytes itself u.ch[0] and u.ch[1] can be occupied, as we are considering the biggest element to b integer, in other words, memory is shared.

So when we see u.i value is equivalent to 512 (the value at 10th bit) +2 (the value at 1st bit) +1 (the value at 0th bit). Therefore 512 + 2 + 1 = 515.