我正在研究 C ++ 11 移动构造函数,但有些东西不起作用。事实上,在我开始编写这样的构造函数之前,问题就出现了。这是一段代码剪辑:
#include <iostream>
#include <string>
#include <sstream>
class Object {
static std::ostream& log(Object &obj) {
std::cout << "Object::id = " << obj.mId << "::";
return std::cout;
}
unsigned mId = 0;
std::string *mText = nullptr;
unsigned nextUniqueId() const {
static unsigned id = 0;
return ++id;
}
const std::string textInfo() const {
std::ostringstream oss;
oss << "mText @ " << &mText;
if (mText) oss << " = " << *mText;
return oss.str();
}
public:
Object() = delete;
Object& operator= (const Object&) = delete;
explicit Object(const std::string& str) : mId(this->nextUniqueId()), mText(new std::string(str)) {
Object::log(*this) << "constructor::one-argument\n";
}
Object(const Object& obj) : mId(this->nextUniqueId()), mText(new std::string(*obj.mText)) {
Object::log(*this) << "constructor::copy\n";
}
virtual ~Object() {
Object::log(*this) << "destructor::" << this->textInfo() << "\n";
if (mText) {
delete mText;
mText = nullptr;
}
}
};
static Object get_object() {
return Object("random text");
}
int main(int argc, char **argv) {
Object a("first object"); // OK
/*
* Expected behaviour: inside get_object() function new Object is created which is then copied into
* variable b. So that new ID should be given.
*/
Object b = get_object(); // What the hell?! Not what expected! Why?
std::cout << std::endl;
return 0;
}
预期输出类似于此:
Object::id = 1::constructor::one-argument
Object::id = 2::constructor::one-argument
Object::id = 2::destructor::mText @ 0x7fff32c25f70 = random text
Object::id = 3::constructor::copy
Object::id = 3::destructor::mText @ <DIFFERENT THAN IN ID=2> = random text
Object::id = 1::destructor::mText @ 0x7fff32c25f90 = first object
我得到了这个:
Object::id = 1::constructor::one-argument
Object::id = 2::constructor::one-argument
Object::id = 2::destructor::mText @ 0x7fff32c25f70 = random text
Object::id = 1::destructor::mText @ 0x7fff32c25f90 = first object
看起来像变量b
的是在现场创建的(可能是inline
之类的东西?)。坦率地说,我不知道发生了什么,有人可以解释一下吗?
答案 0 :(得分:3)
编译器优化了复制/移动全部...
答案 1 :(得分:3)
这称为返回值优化或RVO。编译器选择直接在get_object()
的{{1}}的内存位置创建b
返回的临时值。它受标准认可和非常普遍的优化。
答案 2 :(得分:1)
允许编译器应用“返回值优化”RVO,这就是复制优化的原因。请注意,尽管与输出消息相关的副作用
,标准允许这样做