我有一个小程序,如下所示:
class boovector{
private: int size;
char *arr;
public:
boovector(){size=1;arr=new char[size];cout<<" boovector default constructor called"<<endl;}
boovector(boovector &b)
{
cout<<"boovector copyconstructor called"<<endl;
size = b.size;
arr = new char[size];
strncpy(arr,b.arr,size);
}
boovector(boovector &&b)
{
cout<<"boovector move assignment operator called"<<endl;
size =b.size;
arr = b.arr;
b.arr = nullptr;
}
~boovector()
{
delete []arr;
}
};
boovector createboovector()
{
boovector v;
return v;
}
void foo(boovector v)
{
}
int main(int argc, char *argv[])
{
boovector vet = createboovector();
foo(vet);
foo(createboovector());
return 0;
}
输出
boovector default constructor called
boovector copyconstructor called
boovector default constructor called
我希望在输出中看到名为的"boovector移动赋值操作符“。
如果我注释移动构造函数"boovector(boovector &&b)",我会得到编译器错误
invalid initialization of non-const reference of type 'boovector&' from an
rvalue of type 'boovector'
我想了解move构造函数不被调用背后的逻辑。
发布于 2018-06-30 04:49:33
在我的MSVC 2017社区版上,生成的输出是:
boovector default constructor called
boovector move assignment operator called
boovector copyconstructor called
boovector default constructor called
boovector move assignment operator called
所以它的工作方式和预期的一样。
https://stackoverflow.com/questions/51108324
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