字符串使用C数组的动态分配 [英] Dynamic Allocation of an Array of Strings in C

问题描述

有关家庭作业，我需要定义分配内存以字符串数组（这是为一个结构）的功能。

For home-work, I need to define a function that allocate memory to an array of strings (which is into a struct).

每个字符串的长度被给出：MAX_WORD_LEN + 1（= 10 + 1）的
我不得不为 LEN 串号， LEN 在输入收到分配内存。

The length of each string is given: MAX_WORD_LEN+1 (=10+1) I have to allocate memory for len number of strings, len is recieved in the input.

用绳子定义的数组结构体（给）：

Struct with the array of strings definition (given):

struct dict{
    int len;
    char (*dict0)[MAX_WORD_LEN+1];
    char (*dict1)[MAX_WORD_LEN+1];
};

我不明白的声明字符（* dict0）MAX_WORD_LEN + 1];

• 函数声明也给出：结果
   无效createDict（结构字典* myDict）;

• The function declaration is also given:
  void createDict(struct dict* myDict);

这是我写的，但我不知道它的工作原理，我有困难的时候检查它的编译器。我也写了基于来自网站和其他网站后，并没有真正理解它：

This is what I wrote, but I'm not sure it works, and I have a difficult time checking it in the compiler. I also wrote it based on post from this and other websites and do not really understand it:

选项1：

void createDict(struct dict* myDict)
{

    myDict->(*dict0) =  malloc( (myDict->len)*sizeof(char*));
    myDict->(*dict1) = (char**) malloc( (myDict->len)*sizeof(char*));

    for(int i=0;i<(myDict->len);i++)
    {
        (myDict->(*dict0)[i]) = (char*)malloc((MAX_WORD_LEN+1)*sizeof(char));
        (myDict->(*dict0)[i]) = (char*)malloc((MAX_WORD_LEN+1)*sizeof(char));
    }
}

选项2：

(myDict->(*dict0)[MAX_WORD_LEN+1]) = malloc((myDict->len) * sizeof(char*));
(myDict->(*dict1)[MAX_WORD_LEN+1]) = malloc((myDict->len) * sizeof(char*));

请向我解释...

推荐答案

首先，你需要了解以下三个声明的区别。为简便起见，假设 N 是MAX_WORD_LEN + 1相匹配的大小：

First you need to understand the difference between the following three declarations. For the sake of brevity, assume N is MAX_WORD_LEN+1 to match your sizing:

char data[N];      // array of N chars
char *data[N];     // array of N char *pointers* (i.e. char *)
char (*data)[N];   // pointer to array of N chars

记住高于一切，指针的变量的持有地址，并实现定义。就像一个 INT 变量保存一个实现整数值，指针变量保存的实现地址。

Remember above all else, pointers are variables that hold an "address" and are implementation-defined. Just like an int variable holds the value of an implementation integer, a pointer variable holds an implementation address.

在几乎所有情况下，可以适当的malloc（）内存使用与非关联化的基本目标的sizeof（）运算符指针类型。有些情况下，这是不直观或容易presentable，但下面的内容应该有所帮助：

In almost all cases, you can properly malloc() memory for a pointer type using the sizeof() operator with the underlying target dereferenced. There are some cases where this is not intuitive or easily presentable, but the following should help:

// allocates sizeof(Type) block
Type *p = malloc(sizeof(*p));

// allocates N*sizeof(Type) contiguous blocks
//  note: we'll use this style later to answer your question
Type *pa = malloc(N * sizeof(*pa));

这将工作不管是什么键入是。这一点很重要，因为在你的情况下，你必须声明为指针：

This will work no matter what Type is. This is important, because in your case you have a pointer declared as :

char (*dict)[N];

由于我们已经在上面讨论的，该声明类型的指针（指针对N-字符）。需要注意的是没有实际存储器已被分配。这只是一个指针;仅此而已的。因此，您可以使用上述语法安全分配一个单独的元素：

As we already discussed above, this declares a pointer of type (pointer-to-N-chars). Note that no actual memory has been allocated yet. This is just a pointer; nothing more. Therefore, you can safely allocate a single element using the above syntax as:

// allocate single block
char (*dict)[N] = malloc(sizeof(*dict));

但是，这只占一个条目。您需要 LEN 项，所以：

// allocate 'len' contiguous blocks each N chars in size
char (*dict)[N] = malloc(len * sizeof(*dict));

现在字典被安全地寻址从0阵列。（LEN-1）。你可以在你的数据复制，如：

Now dict is safely addressable as an array from 0..(len-1). You can copy in your data such as:

strcpy(data[0], "test");
strcpy(data[1], "another test");

只要源字符串不超过的N-字符（包括零终止），这将正常工作。

So long as the source string does not exceed N-chars (including the zero-terminator), this will work correctly.

最后，完成后不要忘记释放你分配的：

Finally, don't forget to free your allocation when finished:

free(dict);

---

扰流

myDict->dict0 =  malloc( myDict->len * sizeof(*(myDict->dict0)));
myDict->dict1 =  malloc( myDict->len * sizeof(*(myDict->dict1)));

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