How to Explain this C Union Output

Question

#include <stdio.h>
union p
{
    int x;
    char y;
}
k = {.y = 97};

int main()
{
    printf("%d\n", k.y);
    return 0;
}

OUTPUT: 97

I came across this Question. As we know we can only initialize the first member of Union. But in this, at the time of initialization, the y variable is initialized through some given method!

Can anyone explain this to me how k={ .Y=97} is breaking the rule stated in Dennis Ritchie's book "Union can only be initialized with a value of the type of its first member" and initializing the second variable instead ?

Who told you you can only initialize the first member of some `union`? This is a mistake (you can initialize any single member of some union). — Basile Starynkevitch, Jun 16 '13 at 17:29
The designated initialisers were introduced in c99. dmr did not know about that back in the seventies. — wildplasser, Jun 16 '13 at 17:29

score 15 · Accepted Answer · 2013-06-17T00:05:12.617

15

K&R is a great book, but it is old. In C99 You can do this.

Using a designated initializer in the same example, the following initializes the second union member age :

union {
       char birthday[9];
       int age;
       float weight;
      } people = { .age = 14 };

edited Jun 17 '13 at 00:05

answered Jun 16 '13 at 17:30

You say "K&R ... is old". I tend to think, in the computer world, 35 years (1st edition was 1978) is somewhat beyond "old"... Even the 2nd edition, which IIRC was 1988, is "old"... – twalberg Jul 15 '13 at 19:48

How to Explain this C Union Output

1 Answers1