如何返回 fstream (C++0x)
我想我会直接进入它并从代码开始：
#include <iostream>
#include <fstream>
#include <string>
class test : public std::ofstream
{
public:
test(const std::string& filename) { this->open(gen_filename(filename)); };
test(const test&) = delete;
//test(test&& old) = default; // Didn't compile
test(test&& old) {};
private:
std::string gen_filename(const std::string& filename)
{ return filename + ".tmp"; }
};
int main()
{
auto os = test("testfile");
os << "Test1\n";
os << "Test2\n";
}
基本上，我需要返回一个 ofstream。当然，你不能复制 ofstream，所以我摆弄了类测试中的代码，并且我得到了上面的编译和工作，如你所期望的（在 gcc 4.5 上）。
但我有一种不好的感觉，这只是因为我的编译器在“auto os = test()”上执行了“返回值优化”(RTO)。事实上，如果修改为以下内容：
int main()
{
auto os = test("testfile");
os << "Test1\n";
auto os2 = std::move(os);
os2 << "Test2\n";
}
我不再在输出中同时得到 Test1 和 Test2。
问题是，“test”类不可复制，因此 ofstream 不可能被复制。我只是希望能够从函数返回它。我似乎能够通过 GCC 做到这一点。
我不想取消引用指向分配的流堆的智能指针，或者重新打开文件，因为它当前无需执行这些操作即可工作。我只是觉得我的方法有点“不标准”，所以用标准方法来做我所描述的事情会很棒。
I think I'll get right into it and start with the code:
#include <iostream>
#include <fstream>
#include <string>
class test : public std::ofstream
{
public:
test(const std::string& filename) { this->open(gen_filename(filename)); };
test(const test&) = delete;
//test(test&& old) = default; // Didn't compile
test(test&& old) {};
private:
std::string gen_filename(const std::string& filename)
{ return filename + ".tmp"; }
};
int main()
{
auto os = test("testfile");
os << "Test1\n";
os << "Test2\n";
}
Basically, I need to return an ofstream. Of course you can't copy an ofstream, so I fiddled around with the code in the class test, and I got the above to compile and work as you would expect (on gcc 4.5).
But I have a bad feeling this is just due to my compiler doing "Return Value Optimization" (RTO) on "auto os = test()". Indeed, if modify to the following:
int main()
{
auto os = test("testfile");
os << "Test1\n";
auto os2 = std::move(os);
os2 << "Test2\n";
}
I no longer get both Test1 and Test2 in the output.
The thing is, the class "test" isn't copyable, so there's no chance of the ofstream being duplicated. I just want to be able to return it from a function. And I seem to be able to do that with GCC.
I'd rather not have dereference smart pointers to a heap allocated ofstream, or reopen the file, as it currently works without doing those things. I just have a feeling I'm being a little "non-standard" in my approach, so a standard way of doing what I've described would be great.
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我将在这里回答我自己的问题：
在 GCC C++0x 库中Features 页面，请查看第 27.9 项，其中内容如下：
我想这可能是我在使用 gcc 时遇到的问题。
I'm going to answer my own question here:
In the GCC C++0x Library Features page, have a look at item 27.9, which reads:
I guess that's probably the issue I'm having with gcc.
问题在于：
这可以让你从右值
test构造一个新的test，是的，但是它没有说明你的基础，它只是默认构造的（没有打开文件）。您想要的是这样的：这会将流从
old移动到基础中。The problem is with this:
This lets you construct a new
testfrom an rvaluetest, yes, but it says nothing about your base, which is simply being default constructed (no open file). What you want is this:Which will move the stream from
oldinto the base.调用者是否需要知道您正在返回
ofstream，或者返回streambuf并让调用者将其包装在流中更有意义？Does the caller need to know that you are returning an
ofstream, or would it make more sense to return astreambuf, and let the caller wrap it inside a stream?