Anonymous Union in Struct Initializer

Why does following example not work in C?


#include <stdio.h>

typedef struct {
int x;
} X;

typedef struct {
char y[10];
} Y;

typedef struct {
int pos;
union {
X x;
Y y;
};
} Test;

int main() {
X x = { 65 };
Y y = { "Hallo" };
Test t = { 1, x }; // OK
printf("1: %d %d '%s'\n", t.pos, t.x.x, t.y.y);
Test t2 = { 2, y }; // ERROR
printf("2: %d %d '%s'\n", t2.pos, t2.x.x, t2.y.y);
Test t3 = { 3 }; // OK
printf("3: %d %d '%s'\n", t3.pos, t3.x.x, t3.y.y);
return 0;
}


> main.c: In function ‘main’:<br>
> main.c:25:3: error: incompatible types when initializing type ‘int’ using type ‘Y’<br>
> Test t2 = { 2, y }; // ERROR<br>
> ^


EDIT:
By the way: `t2.y = y;` works

Because the initializer's type isn't analyzed and matched against the possible members of the `union`.

Instead, you're simply supposed to provide an initializer for the first `union` member.

C11 draft §6.7.9.17:

> Each brace-enclosed initializer list has an associated current object. When no designations are present, subobjects of the current object are initialized in order according to the type of the current object: array elements in increasing subscript order, structure members in declaration order, and the first named member of a union.

As mentioned, you can use designators to control this:

Test t2 = { 2, .y = y };

should work (that `.y` is a designator, "new" in C99).

You have to specify that you are initializing a member other than the first:

Test t2 = { 2, { .y = y } } ;

Otherwise the compiler will try to initialize it as if you wrote: `.x = y`