std :: copy / memcpy / memmove优化 [英] std::copy/memcpy/memmove optimizations

问题描述

我查看了GCC STL（4.6.1），并看到 std :: copy（）使用优化版本的情况下内置 __is_trivial（）计算为 true 。

I looked into the GCC STL (4.6.1) and saw that std::copy() uses an optimized version in case the builtin __is_trivial() evaluates to true.

std :: copy（）和 std :: reverse_copy（）模板对于复制数组中的元素非常有用，我想使用它们。但是，我有一些类型（它们是模板实例化的结果），它们是包含一些琐碎值，没有指针，没有复制构造函数或赋值运算符的结构体。

Since the std::copy() and std::reverse_copy() templates are very useful for copying elements in arrays, I would like to use them. However, I have some types (which are results of template instantiations) that are structs that contain some trivial values, no pointers and have no copy constructor or assignment operator.

G ++聪明到足以明白我的类型其实是微不足道的？在C ++ 98中有什么方法来确保STL实现知道我的类型是微不足道的？

Is G++ smart enough to figure out that my type in fact is trivial? Is there any way in C++98 to make sure an STL implementation knows that my type is trivial?

我想在C ++ 11中，事情会变得方便使用 is_trivial<> 类型特征。是这样吗？

I guess that in C++11, things will become convenient using the is_trivial<> type trait. Is this right?

谢谢！

编辑：对不起这么晚，一个非常简单的类型类的示例，它对GCC和llvm不重要。任何想法？

Sorry for being so late with this, but here is an example of a pretty simple Type class that is not trivial to GCC and llvm. Any ideas?

#include <iostream>

struct Spec;

template <typename TValue, typename TSpec>
class Type
{
public:
    TValue value;

    Type() : value(0) {}
};

int main()
{
    std::cerr << "__is_trivial(...) == "
              << __is_trivial(Type<char, Spec>) << '\n';                                                                                                                                                                                                                                    
    return 0;
} 

推荐答案

意味着。

There has been some debate over what trivial meant.

例如，你的例子不可简单地构造， c $ c> std :: is_trivially_default_constructible 会返回false我想）。

For example, your example is not trivially constructible as far as I can tell (std::is_trivially_default_constructible would return false I think).

在你的case，我想你会需要新的trait < a href =http://en.cppreference.com/w/cpp/types/is_trivially_copyable =nofollow> std :: is_trivially_copyable ，这是更细粒度。所以...升级您的编译器？

In your case, I think you would need the new trait std::is_trivially_copyable, which is more fine grained. So... upgrade your compiler ?

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