在C ++中返回对象或指针 [英] Returning an object or a pointer in C++
问题描述
在C ++中，我的方法应该返回对象还是指向对象的指针?如何决定?如果是运营商怎么办?我该如何定义?
In C++, should my method return an object or a pointer to an object? How to decide? What if it's an operator? How can I define?
还有另一件事-如果指针变成向量，返回后如何找出其大小?如果不可能，如我所想，我应该如何继续正确地返回不受此限制的数组?
And one more thing - if the pointer turns to be a vector, how can I find out its size after returned? And if it's impossible, as I think it is, how should I proceed to correctly return an array without this limitation?
推荐答案
在C ++中，我的方法应该返回对象还是指向对象的指针? 如何决定?
In C++, should my method return an object or a pointer to an object? How to decide?
自C ++ 11起，我们就在C ++中使用了移动语义，这意味着它与以前一样容易，现在也很容易按值返回.这应该是默认值.
Since C++11 we have move semantics in C++ which means that it as easy as before and now also fast to return by value. That should be the default.
如果它是运算符怎么办?我该如何定义?
What if it's an operator? How can I define?
许多运算符(例如operator=)通常将引用返回给*this
Many operators such as operator= normally return a reference to *this
X& X::operator=(X rhs);
如果您想遵守通常的模式(并且应该这样做)，则需要为每个操作员查询一下.从此处开始:运算符重载
You need to look that up for each operator if you would like to comply with the usual patterns (and you should). Start here: Operator overloading
Ed S指出，返回值优化也适用(甚至在C ++ 11之前)，这意味着返回的对象经常不需要复制或移动.
As pointed out by Ed S. return value optimization also applies (even before C++11) meaning that often object you return need neither be copied or moved.
所以，现在这是退货的方式:
So, this is now the way to return stuff:
std::string getstring(){
std::string foo("hello");
foo+=" world";
return foo;
}
我在这里制作了一个foo对象这一事实并不是我的意思，即使您只是做return "hello world";这也是要走的路.
The fact that I made a foo object here is not my point, even if you did just do return "hello world"; this is the way to go.
还有另一件事-如果指针变成矢量，我该怎么办 回来后找出它的大小吗?我认为如果不可能的话 是的，我应该如何正确地返回没有这个的数组 限制吗?
And one more thing - if the pointer turns to be a vector, how can I find out its size after returned? And if it's impossible, as I think it is, how should I proceed to correctly return an array without this limitation?
标准中的所有可复制或可移动类型(几乎是所有类型，例如vectors，sets等)都是一样，除了少数例外.例如，std :: arrays不能从移动中受益.它们花费的时间与元素的数量成正比.在那里，您可以将其返回unique_ptr以避免复制.
The same goes for all copyable or movable types in the standard (these are almost all types, for example vectors, sets, and what not), except a few exceptions. For example std::arrays do not gain from moving. They take time proportional to the amount of elements. There you could return it in a unique_ptr to avoid the copy.
typedef std::array<int,15> MyArray;
std::unique_ptr<MyArray> getArray(){
std::unique_ptr<MyArray> someArrayObj(new MyArray());
someArrayObj->at(3)=5;
return someArrayObj;
}
int main(){
auto x=getArray();
std::cout << x->at(3) <<std::endl; // or since we know the index is right: (*x)[3]
}
现在，为避免再写new(在极少数情况下，专家除外)，您应该使用一个名为make_unique的辅助函数.这将极大地帮助异常安全，并且非常方便:
Now, to avoid ever writing new anymore (except for experts in rare cases) you should use a helper function called make_unique. That will vastly help exception safety, and is as convenient:
std::unique_ptr<MyArray> getArray(){
auto someArrayObj=make_unique<MyArray>();
someArrayObj->at(3)=5;
return someArrayObj;
}
要获得更多动力和make_unique的(确实很短)实施，请在此处查看:
make_unique和完美转发
For more motivation and the (really short) implementation of make_unique, have a look here:
make_unique and perfect forwarding
现在make_unique是C ++ 14标准的一部分.如果没有，可以从STL的提案中找到并使用整个实现:
Now make_unique is part of the C++14 standard. If you don't have it, you can find and use the whole implementation from the proposal by S.T.L.:
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