是否可以推断类型是否是不完整的，无编译失败？ [英] Is it possible to deduce whether type is incomplete without compilation failure?

问题描述

我想实现行为像sizeof（complete_type）将返回实际sizeof和sizeof（incomplete_type） - 将只是0

I want to achieve behavior like sizeof(complete_type) will return real sizeof, and sizeof(incomplete_type) - will be just 0

我需要这个提供扩展运行用于具有每种类型描述结构的IPC（进程间）通信的时间类型信息：

I need this to provide extended run time type information for IPC(inter-process) communication with the description structure per type:

struct my_type_info
{
    bool   is_pointer;
    size_t size;         //for double* will be 4 on i386. that is sizeof(double*)
    size_t base_size;    //for double* will be 8. that is sizeof(double)
};

进入我的系统时出现问题类似于MyOnlyDeclaredClass;我收到编译错误，显然是因为我不能把它的大小。

The problem appears when into my system goes something like class MyOnlyDeclaredClass; I got compilation error, obviously by reason I can't take size of it.

boost type_traits http://www.boost.org/doc/libs/1_48_0/libs/type_traits/doc/html/index.html 建议许多编译时类，但没有is_incomplete

boost type_traits http://www.boost.org/doc/libs/1_48_0/libs/type_traits/doc/html/index.html suggests many compile-time classes, but there is no 'is_incomplete'

有趣的编译器是VS2008，VS2010，clang 3，gcc-4.6，gcc-4.7

Interesting compilers are VS2008, VS2010, clang 3, gcc-4.6, gcc-4.7

推荐答案

使用 SFINAE ，一如往常。这是一个可能的实现：

Use SFINAE, as usual. This is one possible implementation:

struct char256 { char x[256]; };

template <typename T>
char256 is_complete_helper(int(*)[sizeof(T)]);

template <typename>
char is_complete_helper(...);

template <typename T>
struct is_complete
{
    enum { value = sizeof(is_complete_helper<T>(0)) != 1 };
};

示例：

#include <cstdio>

struct F;
struct G {};

int main()
{
    printf("%d %d\n", is_complete<F>::value, is_complete<G>::value);
    return 0;
}

（注意： gcc 4.5 （不是因为C ++ 0x）和ang 2.9，但不是gcc 4.3 ）

(Note: Works on gcc 4.5 (no it's not because of C++0x) and clang 2.9, but not gcc 4.3)

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