灵活的数组成员分配结构 [英] Allocating struct with flexible array member

问题描述

这是C99 code：

This is C99 code:

typedef struct expr_t
{
    int n_children; 
    foo data; // Maybe whatever type with unknown alignment
    struct expr_t *children[];
} expr_t;

现在，我怎么分配内存？

Now, how do I allocate memory ?

expr_t *e = malloc (sizeof (expr_t) + n * sizeof (expr_t *));

或

expr_t *e = malloc (offsetof (expr_t, children) + n * sizeof (expr_t *));

是的sizeof 甚至保证与灵活的阵列成员的类型（GCC接受它）？

Is sizeof even guaranteed to work on an type with flexible array member (GCC accepts it) ?

推荐答案

expr_t * E =的malloc（sizeof的（expr_t）+ N * sizeof的（expr_t *））; 很好C99中定义。从C99规范6.7.2.1.16：

expr_t *e = malloc (sizeof (expr_t) + n * sizeof (expr_t *)); is well defined in C99. From the C99 specification 6.7.2.1.16:

作为一个特例，一个结构的多于一个的最后一个元件
  名为构件可以具有一个不完整的数组类型;这就是所谓的
  灵活的数组成员。在大多数情况下，该柔性阵列构件
  被忽略。具体地，结构的尺寸是因为如果
  除了它可能有更灵活的阵列成员已被删去
  尾随填充比遗漏意味着

As a special case, the last element of a structure with more than one named member may have an incomplete array type; this is called a flexible array member. In most situations, the flexible array member is ignored. In particular, the size of the structure is as if the flexible array member were omitted except that it may have more trailing padding than the omission would imply.

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