如何将一个动态数组中在C结构？ [英] How to include a dynamic array INSIDE a struct in C?

问题描述

我环顾四周，但一直未能找到解决什么必须是一个好问的问题。
这里是code我有：

 的#include＆LT;＆stdlib.h中GT;结构my_struct {
    INT N;
    个char []
};诠释的main（）
{
    结构my_struct毫秒;
    ms.s =的malloc（sizeof的（字符*）* 50）;
}
 

和以下是错误的gcc给我：
错误：无效使用灵活的数组成员

我可以得到它来编译，如果我宣布结构内S的声明为

 的char * S
 

这可能是一个卓越的实现（指针运算比数组快，是吗？）
但我认为在C的声明

 个char []
 

是相同的

 的char * S
 

解决方案

您现在写的，曾经被称为结构黑客，直到C99祝福它作为一种柔性数组成员的方式。你得到一个错误（可能是这样）的原因是，它需要跟一个分号：

 的#include＆LT;＆stdlib.h中GT;结构my_struct {
    INT N;
    个char [];
};
 

当你为这个分配空间，要分配的大小结构的以及的要用于阵列的空间量：

 结构my_struct * S =的malloc（sizeof的（结构my_struct）+ 50）;
 

在此情况下，柔性数组成员是char类型的数组，和sizeof（char）的== 1，这样你就不会需要通过其规模倍增，但就像任何其他的malloc你需要，如果这是一些其他类型的数组：

 结构dyn_array {
    INT大小;
    int数据[];
};结构dyn_array my_array =的malloc（sizeof的（结构dyn_array）+ 100 * sizeof的（INT））;
 

编辑：这给出了从该构件改变到指针不同的结果。在这种情况下，（通常）需要两个单独的分配，一个用于结构本身，和一个用于额外的数据要由指针指向。使用灵活的数组成员可以分配在一个块中的所有数据。

I have looked around but have been unable to find a solution to what must be a well asked question. Here is the code I have:

 #include <stdlib.h>

struct my_struct {
    int n;
    char s[]
};

int main()
{
    struct my_struct ms;
    ms.s = malloc(sizeof(char*)*50);
}

and here is the error gcc gives me: error: invalid use of flexible array member

I can get it to compile if i declare the declaration of s inside the struct to be

char* s

and this is probably a superior implementation (pointer arithmetic is faster than arrays, yes?) but I thought in c a declaration of

char s[]

is the same as

char* s

解决方案

The way you have it written now , used to be called the "struct hack", until C99 blessed it as a "flexible array member". The reason you're getting an error (probably anyway) is that it needs to be followed by a semicolon:

#include <stdlib.h>

struct my_struct {
    int n;
    char s[];
};

When you allocate space for this, you want to allocate the size of the struct plus the amount of space you want for the array:

struct my_struct *s = malloc(sizeof(struct my_struct) + 50);

In this case, the flexible array member is an array of char, and sizeof(char)==1, so you don't need to multiply by its size, but just like any other malloc you'd need to if it was an array of some other type:

struct dyn_array { 
    int size;
    int data[];
};

struct dyn_array my_array = malloc(sizeof(struct dyn_array) + 100 * sizeof(int));

Edit: This gives a different result from changing the member to a pointer. In that case, you (normally) need two separate allocations, one for the struct itself, and one for the "extra" data to be pointed to by the pointer. Using a flexible array member you can allocate all the data in a single block.

这篇关于如何将一个动态数组中在C结构？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！