问题描述
我们可以声明一个可变长度的结构元素吗?
Can we declare a structure element of variable length?
条件如下:
typedef struct
{
uint8_t No_Of_Employees;
uint8_t Employee_Names[No_Of_Employees][15];
}st_employees;
推荐答案
如果在 C99 中编码或 C11,您可能想要使用 灵活数组成员(您没有给出明确的维度,但你应该在头脑中在运行时有一个关于它的约定).
If coding in C99 or C11, you might want to use flexible array members (you don't give an explicit dimension, but you should have a convention about it at runtime in your head).
typedef struct {
unsigned No_Of_Employees;
char* Employee_Names[]; // conventionally with No_of_Employees slots
}st_employees;
对于任何数组,灵活数组成员的每个插槽都有固定的大小.我正在使用指针(例如,我的 Linux/x86-64 机器上的 8 个字节).
As for any array, each slot of a flexible array member has a fixed size. I'm using a pointer (e.g. 8 bytes on my Linux/x86-64 machine).
然后您将使用例如分配这样的结构
Then you would allocate such a structure using e.g.
st_employees* make_employees(unsigned n) {
st_employees* s = malloc(sizeof(s_employees)+n*sizeof(char*));
if (!s) { perror("malloc make_employees"); exit(EXIT_FAILURE); };
s->No_of_Employees = n;
for (unsigned i=0; i<n; i++) s->Employe_Names[i] = NULL;
return s;
}
你可能会使用 (with strdup(3)在堆中复制一个字符串)就像
and you might use (with strdup(3) duplicating a string in the heap) it like
st_employees* p = make_employees(3);
p->Employee_Names[0] = strdup("John");
p->Employee_Names[1] = strdup("Elizabeth");
p->Employee_Names[2] = strdup("Brian Kernighan");
您需要一个 void destroy_employee(st_employee*e)
函数(留给读者作为练习).它可能应该循环 i
到 free
每个 e->Employee_Names[i]
,然后 free(e);...
You'll need a
void destroy_employee(st_employee*e)
function (left as an exercise to the reader). It probably should loop on i
to free
every e->Employee_Names[i]
, then free(e);
...
不要忘记记录有关内存使用的约定(谁负责调用
malloc
和 free
).阅读有关 C 动态内存分配的更多信息(并且害怕 内存碎片 和 缓冲区溢出 和任何其他未定义行为).
Don't forget to document the conventions about memory usage (who is in charge of calling
malloc
and free
). Read more about C dynamic memory allocation (and be scared of memory fragmentation and buffer overflows and any other undefined behavior).
如果使用 GCC 早于 GCC 5 一定要使用
gcc -std=c99 -Wall
编译,因为旧的 GCC 4 编译器的默认标准是 C89.对于较新的编译器,请询问所有警告以及更多警告,例如gcc -Wall -Wextra
...
If using a GCC older than GCC 5 be sure to compile with
gcc -std=c99 -Wall
since the default standard for old GCC 4 compilers is C89. For newer compilers, ask for all warnings and more of them, e.g. gcc -Wall -Wextra
...
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