是C-结构与保证在内存相同的布局相同的成员类型？ [英] Are C-structs with the same members types guaranteed to have the same layout in memory?

问题描述

从本质上讲，如果我有

  typedef结构{
    INT X;
    诠释Ÿ;
} 一个;typedef结构{
    INT H;
    时int k;
} B：
 

和我有 A A ，请问C标准保证（（B *）及一） - GT; k 相同唉？

解决方案

  

是C-结构与保证在内存相同的布局相同的成员类型？

几乎是肯定的。对我来说足够接近。

从n1516，第6.5.2.3条第6款：

  

...如果联合包含共享一个共同的初始序列...几种结构，如果联合对象当前包含这些结构中的一个，它被允许在任何地方，一个检查的任何它们的共同初始部分完成型工会的声明是可见的。两种结构共享的公共初始序列如果相应成员具有兼容的类型（并且，对于位字段，相同的宽度）为一个或多个初始成员的序列

这意味着，如果你有以下的code：

 结构A {
    INT X;
    诠释Ÿ;
};结构B {
    INT H;
    时int k;
};工会{
    结构一个一个;
    结构B B;
} U;
 

如果您分配给 u.a ，标准说，你可以从 u.b 读出相应的值。它横跨合理性的界限表明结构A 和结构b 可以有不同的布局，考虑这一要求。这样的系统将在极端病态的。

记住，标准还保证：

• 结构永远不会重陷presentations。




• 在一个结构中增加字段（斧总是前唉）的地址。




• 第一个字段的偏移始终为零。

然而，这是非常重要的<！/ H1>

您转述的问题，

  

做C标准保证（（B *）及一） - GT;·K 相同a.y

没有！而且它很明确地指出，他们是不一样的！

 结构A {INT X; };
结构B {INT X; };
INT测试（int值）
{
    结构一个一个;
    a.x =价值;
    返回（（结构的b *）及一） -  X的催化剂;
}
 

这是一个别名冲突。

Essentially, if I have

typedef struct {
    int x;
    int y;
} A;

typedef struct {
    int h;
    int k;
} B;

and I have A a, does the C standard guarantee that ((B*)&a)->k is the same as a.y?

解决方案

Are C-structs with the same members types guaranteed to have the same layout in memory?

Almost yes. Close enough for me.

From n1516, Section 6.5.2.3, paragraph 6:

... if a union contains several structures that share a common initial sequence ..., and if the union object currently contains one of these structures, it is permitted to inspect the common initial part of any of them anywhere that a declaration of the completed type of the union is visible. Two structures share a common initial sequence if corresponding members have compatible types (and, for bit-fields, the same widths) for a sequence of one or more initial members.

This means that if you have the following code:

struct a {
    int x;
    int y;
};

struct b {
    int h;
    int k;
};

union {
    struct a a;
    struct b b;
} u;

If you assign to u.a, the standard says that you can read the corresponding values from u.b. It stretches the bounds of plausibility to suggest that struct a and struct b can have different layout, given this requirement. Such a system would be pathological in the extreme.

Remember that the standard also guarantees that:

• Structures are never trap representations.

• Addresses of fields in a structure increase (a.x is always before a.y).

• The offset of the first field is always zero.

However, and this is important!

You rephrased the question,

does the C standard guarantee that ((B*)&a)->k is the same as a.y?

No! And it very explicitly states that they are not the same!

struct a { int x; };
struct b { int x; };
int test(int value)
{
    struct a a;
    a.x = value;
    return ((struct b *) &a)->x;
}

This is an aliasing violation.

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