outputCode:main() { union u { struct s{int a;int b}n; struct ss{int c;long d}ni; }uu; uu.ni.c=1; uu.ni.d=0; printf("%d %d",(uu.n.a),(uu.n.b)); }
1
32767
Now my question is that even though I put d=0 then why does it shows 32767 as output for b?As per my knowledge union members share the same place of memory..Then why is this hapenning here?
When you assign a value for one member of a union, you override the value(s) stored in the other member(s). The other values are no longer valid.
Yeah that i know..in my case it means wirting to c and d means overiding the values of a and b...But why does that 32767 come??
It is probably a consequence of your compiler using ones-complement representation of signed integral types.
I think that its because of padding that extra bits are added..Because if i do sizeof(struct ss) it gives 16 and not 12...Please contribute if I am wrong?Originally Posted by grumpy
Possibly. Possibly not.
Whatever the cause, your code exhibits undefined behaviour. Any results you get are therefore technically correct, according to the standard. If someone found that your code triggered segmentation faults, that would also be technically correct. The reasons come down to how your compiler is implemented.
In a C forum, the actual reason for the behaviour does not matter. To a practical C developer, the only conclusion of interest is that they should not be using unions in the manner that you are.
@grumpy thank you for your valuable contribution..
> As per my knowledge union members share the same place of memory
By the time you've taking into padding and alignment, the amount of overlap between any two apparently adjacent variables might be zero.
Which is seems to be in this case.
The assignment of ni.d completely misses all the bytes where n.b would reside, so all you see is random garbage.Code:#include <stdio.h> #include <stdlib.h> #include <stddef.h> int main() { union u { struct s { int a; int b; } n; struct ss { int c; long d; } ni; } uu; uu.ni.c = 1; uu.ni.d = 0; printf("%d %d\n", (uu.n.a), (uu.n.b)); printf("n.a is at %ld\nn.b is at %ld\nni.d is at %ld\n", offsetof(union u,n.a), offsetof(union u,n.b), offsetof(union u,ni.d) ); return 0; } $ ./a.out 1 32767 n.a is at 0 n.b is at 4 ni.d is at 8
The a and c members, by virtue of being the first member in their respective structs, AND of the same type, can be freely assigned in one and inspected in the other with no problem at all.
But this is a unique special case.
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
