In which manner the value of union members got changed? Any one pls!

Question

https://code.sololearn.com/c351sX8EGY9A/?ref=app

Answer 1

<Saurabh Singh Rawat/> 👋
actually the first 8 bits here (ch[1]), in the memory are the last, and so when the machine read it its read the first 8 bits from right to left (or to precise much more from the low address up to the highest) :v
for example :
char str[]={'a', 0,0,0} ; means :
._24b__|___8b____ #b=bit
00...000|01100001
str[3:1] | str[0]
and as we know the address of str[0] less than str[1,2 or 3]
so if we do something like this :
int *p = (int*) str;
int a=*p ; now the value stored in a is 97 ('a' = 97 ASCII table)
.. anyway there is a lot of things i want to tell you about it🥰 but.. i suggest to you to read about memory and stack and how they work and stuff like that 👌.. and if you want some help, just ask 😄

Answer 2

nothing is wrong ,, first you initialize u.i by 2 and then you initialize the array u.ch :
3=11 (in binary representation)
2=10 (in b..)
and as we know changing one element from a union variable means changing the first sizeof(element) bytes starting from the add of that element..
and we know that sizeof(short) = sizeof(ch) =2 so when you get the value stored at u.i actually means that you get the hole value in u (cuz sizeof(u) = sizeof(u.i)):
it's look like this in memory:
u.i=2
|00000000|00000010
at this point if you check out the value stored at ch[0] you will find it = 2 and ch[1] =0.
now:
ch[0]=3;ch[1]=2;

ch[1] ch[0]
|00000010|00000011=515

Answer 3

Wow! Great explanation.
Thank you so much. :)

Answer 4

Amine Laaboudi could you pls tell me one more thing why first 8 bit are allotted to index 1 not 0 here?

Answer 5

Amine Laaboudi
Wow, you explained again well.
And yes of course I will read about memory and stack, and if I get any problem, will definitely ask you.
You're Awesome!
Thanks again.

Answer 6

<Saurabh Singh Rawat/> you are welcome bro😁