C语言 为结构数组动态分配内存
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/19948733/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Dynamically allocate memory for Array of Structs
提问by Mark L?we
Here's what I'm trying to do:
这是我想要做的：
#include <stdio.h>
#include <stdlib.h>
struct myStruct {
int myVar;
}
struct myStruct myBigList = null;
void defineMyList(struct myStruct *myArray)
{
myStruct *myArray = malloc(10 * sizeof(myStruct));
*myArray[0] = '42';
}
int main()
{
defineMyList(&myBigList);
}
I'm writing a simple C program to accomplish this. I'm using the GNU99 Xcode 5.0.1 compiler. I've read many examples, and the compiler seems to disagree about where to use the structtag. Using a structreference inside the sizeof()command doesn't seem to recognize the structat all.
我正在编写一个简单的 C 程序来实现这一点。我正在使用 GNU99 Xcode 5.0.1 编译器。我读过很多例子，编译器似乎不同意在哪里使用struct标签。struct在sizeof()命令中使用引用似乎根本无法识别struct。
回答by unwind
There are a few errors in your code. Make it:
您的代码中有一些错误。做了：
struct myStruct *myBigList = NULL; /* Pointer, and upper-case NULL in C. */
/* Must accept pointer to pointer to change caller's variable. */
void defineMyList(struct myStruct **myArray)
{
/* Avoid repeating the type name in sizeof. */
*myArray = malloc(10 * sizeof **myArray);
/* Access was wrong, must use member name inside structure. */
(*myArray)[0].myVar = 42;
}
int main()
{
defineMyList(&myBigList);
return 0; /* added missing return */
}
Basically you must use the structkeyword unless you typedefit away, and the global variable myBigListhad the wrong type.
基本上你必须使用struct关键字，除非你typedef离开它，并且全局变量myBigList的类型错误。
回答by fkl
This is because struct name is not automatically converted into a type name. In C (not C++) you have to explicitly typedef a type name.
这是因为struct name 不会自动转换为类型 name。在 C（不是 C++）中，你必须显式 typedef 一个类型名称。
Either use
要么使用
struct myStruct instance;
when using the type name OR typedef it like this
像这样使用类型名称或 typedef 时
typedef struct {
int myVar;
} myStruct;
now myStructcan simply be used as a type name similar to int or any other type.
现在myStruct可以简单地用作类似于 int 或任何其他类型的类型名称。
Note that this is only needed in C. C++ automatically typedefs each struct / class name.
请注意，这仅在 C 中需要。C++ 自动对每个结构/类名称进行类型定义。
A good convention when extending this to structs containing pointers to the same type is here
将其扩展到包含指向同一类型的指针的结构时，一个很好的约定是here
回答by x5lcfd
sizeof(struct myStruct)
or
或者
typedef struct myStruct myStrut;
sizeof(myStruct)
回答by RazvanD
In order to work for all 10 elements for that array the line:
为了处理该数组的所有 10 个元素，该行：
myArray[0].myVar = '42';
should be:
应该：
(*myArray)[0].myVar = '42';
回答by Nirav Ambaliya
Shouldn't the following statement
不应该是下面的说法
myArray[0].myVar = '42';
be this?
是这个？
(*myArray)[0].myVar = 42;
myvar is an integer.
myvar 是一个整数。
