相关问题:
我发布这个问题是因为,这种移动语义确实让我感到困惑。最初,它们对我来说似乎很清楚,但是当我尝试向自己演示这些功能的使用时,我意识到也许我误解了一些东西。
我试图通过移动语义将以下文件安排为类似于矢量的类的简单实现(实际上,main
函数也在那里,还有一个免费的功能使打印到屏幕更容易...)。这并不是真正的 minimum 工作示例,但是它在屏幕上产生的输出是相当可读的,恕我直言。
不过,如果您觉得最好将其缩小,请建议我该怎么做。
无论如何下面的代码,
#include<iostream>
using namespace std;
int counter = 0; // to keep count of the created objects
class X {
private:
int id = 0; // hopefully unique identifyier
int n = 0;
int * p;
public:
// special member functions (ctor, dtor, ...)
X() : id(counter++), n(0), p(NULL) { cout << "default ctor (id " << id << ")\n"; }
X(int n) : id(counter++), n(n), p(new int[n]) { cout << "param ctor (id " << id << ")\n"; };
X(const X& x) : id(counter++), n(x.n), p(new int[n]) {
cout << "copy ctor (id " << id << ") (allocating and copying " << n << " ints)\n";
for (int i = 0; i < n; ++i) {
p[i] = x.p[i];
}
};
X(X&& x) : id(counter++), n(x.n), p(x.p) {
cout << "move ctor (id " << id << ")\n";
x.p = NULL;
x.n = 0;
};
X& operator=(const X& x) {
cout << "copy assignment (";
if (n < x.size() && n > 0) {
cout << "deleting, ";
delete [] p;
n = 0;
}
if (n == 0) {
cout << "allocating, and ";
p = new int[n];
}
n = x.size();
cout << "copying " << n << " values)";
for (int i = 0; i < n; ++i) {
p[i] = x.p[i];
}
cout << endl;
return *this;
};
X& operator=(X&& x) {
this->n = x.n;
this->p = x.p;
x.p = NULL;
x.n = 0;
cout << "move assignment (\"moving\" " << this->n << " values)\n";
return *this;
};
~X() {
cout << "dtor on id " << id << " (array of size " << n << ": " << *this << ")\n";
delete [] p;
n = 0;
}
// getters/setters
int size() const { return n; }
// operators
int& operator[](int i) const { return p[i]; };
X operator+(const X& x2) const {
cout << "operator+\n";
int n = min(x2.size(), this->size());
X t(n);
for (int i = 0; i < n; ++i) {
t.p[i] = this->p[i] + x2.p[i];
}
return t;
};
// friend function to slim down the cout lines
friend ostream& operator<<(ostream&, const X&);
};
int main() {
X x0;
X x1(5);
X x2(5);
x1[2] = 3;
x2[3] = 4;
cout << "\nx0 = x1 + x2;\n";
x0 = x1 + x2;
cout << "\nX x4(x1 + x2);\n";
X x4(x1 + x2);
cout << x4 << endl;
cout << '\n';
}
// function to slim down the cout lines
ostream& operator<<(ostream& os, const X& x) {
os << '[';
for (int i = 0; i < x.size() - 1; ++i) {
os << x.p[i] << ',';
}
if (x.size() > 0) {
os << x.p[x.size() - 1];
}
return os << ']';
}
当我编译并运行它时
$ clear && g++ moves.cpp && ./a.out
输出如下(手动添加#
条评论)
default ctor (id 0)
param ctor (id 1)
param ctor (id 2)
x0 = x1 + x2;
operator+
param ctor (id 3)
move assignment ("moving" 5 values)
dtor on id 3 (array of size 0: [])
X x4(x1 + x2);
operator+
param ctor (id 4)
[0,0,3,4,0]
dtor on id 4 (array of size 5: [0,0,3,4,0])
dtor on id 2 (array of size 5: [0,0,0,4,0])
dtor on id 1 (array of size 5: [0,0,3,0,0])
dtor on id 0 (array of size 5: [0,0,3,4,0])
从输出的第一部分开始,我想我确实演示了移动分配运算符的预期用法。在这方面我是对的吗? (从下一个输出中,我似乎不是,但是我不确定。)
在这一点上,如果我认为复制省略阻止了对复制ctor的调用是正确的,那么我自然会想到一个问题(and not only me, see OP's comment here):
不是基于x4
中x1 + x2
的结果基于另一个临时对象(例如 X x4(x1 + x2);
创建对象的情况)正是应该为其引入移动语义的一种?如果不是,那么显示移动ctor用法的基本示例是什么?
然后我读到可以通过添加适当的选项来防止复制省略。
的输出
clear && g++ -fno-elide-constructors moves.cpp && ./a.out
但是,以下内容是
:default ctor (id 0)
param ctor (id 1)
param ctor (id 2)
x0 = x1 + x2;
operator+
param ctor (id 3)
move ctor (id 4)
dtor on id 3 (array of size 0: [])
move assignment ("moving" 5 values)
dtor on id 4 (array of size 0: [])
X x4(x1 + x2);
operator+
param ctor (id 5)
move ctor (id 6)
dtor on id 5 (array of size 0: [])
move ctor (id 7)
dtor on id 6 (array of size 0: [])
[0,0,3,4,0]
dtor on id 7 (array of size 5: [0,0,3,4,0])
dtor on id 2 (array of size 5: [0,0,0,4,0])
dtor on id 1 (array of size 5: [0,0,3,0,0])
dtor on id 0 (array of size 5: [0,0,3,4,0])
+enrico:CSGuild$
现在看起来像是我期望的对移动ctor的调用,但是该调用和对移动分配的调用都在前面分别有另一个对移动ctor的调用。
为什么会这样?我是否完全误解了移动语义的含义?
答案 0 :(得分:3)
您似乎在这里有两个问题:
X x4(x1 + x2)
调用move构造函数? 不是那种情况(X x4(x1 + x2);)正是这种情况 应该在哪里引入语义?
嗯,不。为了使用移动语义,您实际上建议我们选择在X
中构造一个operator+
,然后将其移动到结果{{1 }} ,相比之下,复制复制的构造x4
期间就位的最终结果(x4
)效率低下。
已禁用复制删除功能,为什么在operator+
期间看到两次对move构造函数的调用?考虑到这里有三个作用域:
X x4(x1 + x2)
范围,我们在其中构造operator+
并将其返回; X
的{{1}}范围; main
构建X x4(x1 + x2)
的{{1}}; 然后,在没有省略的情况下,编译器为:
X constructor
移至X
(移至x1 + x2
);和operator+
的内容移至main
。