Anonymous union in initializer

Question

Why doesn't the following example 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, txx, tyy); Test t2 = { 2, y }; // ERROR printf("2: %d %d '%s'\n", t2.pos, t2.xx, t2.yy); Test t3 = { 3 }; // OK printf("3: %d %d '%s'\n", t3.pos, t3.xx, t3.yy); return 0; } 

main.c: In the function 'main:
main.c: 25: 3: error: incompatible types when initializing the type "int using type" Y
Testing t2 = {2, y}; // ERROR
^

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

Answer 1

Since the type of initializer is not parsed and does not match against possible union members.

Instead, you just have to provide an initializer for the first member of union .

Draft C11 §6.7.9.17:

Each initializer list, enclosed in braces, has a current object associated with it. When there is no designation, the subobjects of the current object are initialized in the order corresponding to the type of the current object: array elements in ascending index order, structure members in the declaration order, and the first named member of the union.

As already mentioned, you can use pointers to control this:

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

should work (which .y is the designation "new" on C99).

Answer 2

You must indicate that you are initializing an element other than the first:

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

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