C 编程 - "derived"类型的基本结构类型

Question

这个想法是可以创建一个指向 Struct_A 实例的指针或
带有 BaseStruct 类型指针的 Struct_B 实例。

我可以使用任何编译器在 C 中创建“基本结构类型”吗？
我用 GCC 测试了它并且它有效，但我不确定它是否在任何地方都有效......

#include <stdio.h>

typedef enum {
 StructType_A,
 StructType_B
} StructType;

// >>> base type <<<
typedef struct {
 StructType type;
} BaseStruct;

// >>> type A <<<
typedef struct {
 StructType type;
 int xyz;
 char blabliblub;
} Struct_A;

// >>> type B <<<
typedef struct {
 StructType type;
 long int abc;
} Struct_B;

int main () {
 Struct_A a;
 BaseStruct* ptr;

 a.type = StructType_A;
 a.xyz  = 7853;
 a.blabliblub = 'z';

 ptr = (BaseStruct*) &a;

 if (ptr->type == StructType_A) printf ("YAY :3\n");
}

Answer 1

首先，回答你的问题，不允许用BaseStruct类型的指针指向Struct_B类型，如果这两种类型有不同的对齐要求(参见C standard):

6.3.2.3 Pointers

(7) A pointer to an object or incomplete type may be converted to a pointer to a different object or incomplete type. If the resulting pointer is not correctly aligned for the pointed-to type, the behavior is undefined.

但是，您可以通过建立适当的联盟轻松克服这种情况:

typedef enum {
    StructType_A,
    StructType_B
} StructType;

// >>> type A <<<
typedef struct {
    int xyz;
    char blabliblub;
} Struct_A;

// >>> type B <<<
typedef struct {
    long int abc;
} Struct_B;

// >>> composite type <<<
typedef struct {
    StructType type;
    union {
        Struct_A a;
        Struct_B b;
    } content;
} Union_AB;


int main () {
    Struct_A a;
    a.xyz  = 7853;
    a.blabliblub = 'z';

    Union_AB anA;
    anA.type = StructType_A;
    anA.content.a.xyz = 7853;
    anA.content.a.blabliblub = 'Z';

    Union_AB someB;
    someB.type = StructType_B;
    someB.content.b.abc = 5653L;

    Union_AB* ptr = (&anA);
    if (ptr->type == StructType_A) printf ("YAY :3\n");

    ptr = (&someB);
    if (ptr->type == StructType_A) printf ("YAY :3\n");

    return 0;
}