是否可以将指针从结构类型转换为另一种扩展 C 中第一个的结构类型? [英] Is it possible to cast pointers from a structure type to another structure type extending the first in C?

问题描述

如果我有结构定义，例如，像这样:

If I have structure definitions, for example, like these:

struct Base {
  int foo;
};

struct Derived {
  int foo; // int foo is common for both definitions
  char *bar;
};

我可以做这样的事情吗?

Can I do something like this?

void foobar(void *ptr) {
  ((struct Base *)ptr)->foo = 1;
}

struct Derived s;

foobar(&s);

换句话说，当它的类型实际上是 Derived * 时，我可以将 void 指针转换为 Base * 以访问它的 foo 成员吗?

In other words, can I cast the void pointer to Base * to access its foo member when its type is actually Derived *?

推荐答案

许多现实世界的 C 程序假定您展示的构造是安全的，并且有对 C 标准的解释(特别是公共初始序列"规则，C99 §6.5.2.3 p5) 符合该规则.不幸的是，在我最初回答这个问题后的五年里，我可以轻松获得的所有编译器(即 GCC 和 Clang)都收敛于对共同初始序列规则的不同的、更窄的解释，在该规则下，您展示的构造引发未定义的行为.具体来说，试验这个程序:

Many real-world C programs assume the construct you show is safe, and there is an interpretation of the C standard (specifically, of the "common initial sequence" rule, C99 §6.5.2.3 p5) under which it is conforming. Unfortunately, in the five years since I originally answered this question, all the compilers I can easily get at (viz. GCC and Clang) have converged on a different, narrower interpretation of the common initial sequence rule, under which the construct you show provokes undefined behavior. Concretely, experiment with this program:

#include <stdio.h>
#include <string.h>

typedef struct A { int x; int y; }          A;
typedef struct B { int x; int y; float z; } B;
typedef struct C { A a;          float z; } C;

int testAB(A *a, B *b)
{
  b->x = 1;
  a->x = 2;
  return b->x;
}

int testAC(A *a, C *c)
{
  c->a.x = 1;
  a->x = 2;
  return c->a.x;
}

int main(void)
{
  B bee;
  C cee;
  int r;

  memset(&bee, 0, sizeof bee);
  memset(&cee, 0, sizeof cee);

  r = testAB((A *)&bee, &bee);
  printf("testAB: r=%d bee.x=%d
", r, bee.x);

  r = testAC(&cee.a, &cee);
  printf("testAC: r=%d cee.x=%d
", r, cee.a.x);

  return 0;
}

在启用优化(并且没有 -fno-strict-aliasing)的情况下编译时，GCC 和 Clang 都将假定 testAB 的两个指针参数不能指向同一个对象，所以我得到像

When compiling with optimization enabled (and without -fno-strict-aliasing), both GCC and Clang will assume that the two pointer arguments to testAB cannot point to the same object, so I get output like

testAB: r=1 bee.x=2
testAC: r=2 cee.x=2

他们没有对 testAC 做出这样的假设，但是 - 之前的印象是 testAB 需要被编译，就好像它的两个参数可以 指向同一个对象——我对自己对标准的理解不再有足够的信心来判断 那个 是否可以保证继续工作.

They do not make that assumption for testAC, but — having previously been under the impression that testAB was required to be compiled as if its two arguments could point to the same object — I am no longer confident enough in my own understanding of the standard to say whether or not that is guaranteed to keep working.

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