c - C 中的 "Dynamic Inheritance"

Question

我写了下面的代码并且它有效，但我想知道是否可以确保它在所有 x86 机器上一直有效。

#include <stdio.h>
#include <stdlib.h>

typedef struct Base {
  int a;
  float b;
} Base;

typedef struct Derived1 {
  int a;      // This two members have the same position as in the Base
  float b;

  // adding some other members to this struct
  int otherMember;
  int otherMember2;
} Derived1;

int main()
{
  Base *bases[2];

  // Filling the array with different structs
  bases[0] = (Base*) malloc(sizeof(Base));
  bases[1] = (Base*) malloc(sizeof(Derived1));

  bases[1]->a = 5;
  Derived1 *d1 = (Derived1*) bases[1];

  if(d1->a == 5)
    printf("SUCCESS\n");

  return 0;
}

我知道为什么这个例子有效，但它总是有效吗？是否有填充或类似的东西可以阻止它工作，或者 C 标准是否支持它？

Answer 1

根据 C99 规则，这两个 struct 是不兼容的:

two structure, union, or enumerated types declared in separate translation units are compatible if their tags and members satisfy the following requirements: If one is declared with a tag, the other shall be declared with the same tag. If both are complete types, then the following additional requirements apply: there shall be a one-to-one correspondence between their members such that each pair of corresponding members are declared with compatible types, and such that if one member of a corresponding pair is declared with a name, the other member is declared with the same name. For two structures, corresponding members shall be declared in the same order.

您的代码打破了成员之间的一对一对应关系，因此根据标准，这将是无效的:

Base *d1 = (Base*) bases[1];
d1->a=5; // Not valid

幸运的是，您可以通过将 Base 嵌入到 Derived1 中轻松使其有效:

typedef struct Derived1 {
  Base base;
  // adding some other members to this struct
  int otherMember;
  int otherMember2;
} Derived1;

根据 C99，

A pointer to a structure object, suitably converted, points to its initial member

因此，这是有效的:

Base *d1 = (Base*) bases[1];
d1->a=5; // Valid

注意: This Q&A谈论严格别名的相关主题。