使用未指定大小、空括号定义的静态数组?

2022-01-17 00:00:00 class compiler-construction c++ static-array

对于下面的 C++ 代码片段:

For the C++ code fragment below:

class Foo {
    int a[]; // no error
};

int a[];     // error: storage size of 'a' isn't known

void bar() {
    int a[]; // error: storage size of 'a' isn't known
}

为什么成员变量也不会导致错误?这个成员变量是什么意思?

why isn't the member variable causing an error too? and what is the meaning of this member variable?

我通过 CodeBlocks 8.02 使用 gcc 版本 3.4.5(mingw-vista special).

I'm using gcc version 3.4.5 (mingw-vista special) through CodeBlocks 8.02.

在 Visual Studio Express 2008 - Microsoft(R) C/C++ Optimizing Compiler 15.00.30729.01 for 80x86 上，我收到以下消息:

On Visual Studio Express 2008 - Microsoft(R) C/C++ Optimizing Compiler 15.00.30729.01 for 80x86, I got the following messages:

class Foo {
    int a[]; // warning C4200: nonstandard extension used : zero-sized array in struct/union - Cannot generate copy-ctor or copy-assignment operator when UDT contains a zero-sized array
};

int a[];

void bar() {
    int a[]; // error C2133: 'a' : unknown size
}

现在，这也需要一些解释.

Now, this needs some explaination too.

推荐答案

C99 支持称为灵活"数组成员的东西，它允许成为结构的最后一个成员.当您动态分配这样的结构时，您可以增加从 malloc() 请求的数量，以便为数组提供内存.

C99 supports something called a 'flexible' array member that is allowed to be the last member of a struct. When you dynamically allocate such a struct you can increase the amount requested from malloc() to provide for memory for the array.

一些编译器将此作为 C90 和/或 C++ 的扩展添加.

Some compilers add this as an extension to C90 and/or C++.

所以你可以有如下代码:

So you can have code like the following:

struct foo_t {
    int x;
    char buf[];
};


void use_foo(size_t bufSize)
{
    struct foo_t* p = malloc( sizeof( struct foo_t) + bufSize);

    int i;

    for (i = 0; i < bufSize; ++i) {
        p->buf[i] = i;
    }
}

您不能直接定义具有灵活数组成员的结构(作为本地或全局/静态变量)，因为编译器不知道要为其分配多少内存.

You can't define a struct with a flexible array member directly (as a local or a global/static variable) as the compiler won't know how much memory to allocate for it.

老实说，我不确定您如何使用 C++ 的 new 运算符轻松使用这样的东西 - 我认为您必须使用 malloc() 为对象分配内存 并使用放置 new.也许可以使用一些特定于类/结构的 operator new 重载...

I'm honestly not sure how you'd easily use such a thing with C++'s new operator - I think you'd have to allocate the memory for the object using malloc() and use placement new. Maybe some class/struct specific overload of operator new could be used...

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