结构初始化器中的匿名联合

Question

为什么以下示例在 C 中不起作用？

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

编辑：顺便说一句：t2.y = y; 有效

Answer 1

因为未针对 union.

的可能成员分析和匹配初始化程序的类型

相反，您只需为第一个 union 成员提供一个初始化程序。

C11 草案 §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.

如前所述，您可以使用指示符来控制它：

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

应该可以工作（.y 是一个指示符，在 C99 中是 "new"）。

Answer 2

您必须指定要初始化的成员不是第一个：

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

否则编译器将尝试初始化它，就好像您这样写：.x = y

结构初始化器中的匿名联合

Anonymous Union in Struct Initializer

c

struct

initialization

unions