为什么这个复制构造函数而不是移动构造函数? [英] Why is this copy constructor called rather than the move constructor?
问题描述
以下代码段导致在我希望调用move构造函数的地方调用复制构造函数:
#include < cstdio>
struct Foo
{
Foo(){puts(Foo gets built! }
Foo(const Foo& foo){puts(Foo gets copied! }
Foo(Foo& foo){puts(Foo gets moved!); }
};
struct Bar {Foo foo; };
Bar Meow(){Bar bar;返回栏}
int main(){Bar bar(Meow()); }在VS11测试版,在调试模式下,这打印:
$ b < b $ b Foo构建!
Foo被复制!
Foo被复制!
我检查了标准,并且 Bar
满足所有的要求,有一个默认的移动构造函数自动生成,但似乎没有发生,除非有另一个原因,为什么对象不能移动。我在这里看到了很多关于移动和复制构造函数的问题,但我不认为任何人都有这个具体问题。
任何指向这里发生的事?这是标准的行为吗?
解决方案不幸的是,VS11不提供默认的移动构造函数。请参阅备注部分中的移动语义 - 引用: / p>
与默认的复制构造函数不同,编译器不提供
默认的移动构造函数。
The following code snippet causes the copy constructor to be called where I expected the move constructor to be called:
#include <cstdio>
struct Foo
{
Foo() { puts("Foo gets built!"); }
Foo(const Foo& foo) { puts("Foo gets copied!"); }
Foo(Foo&& foo) { puts("Foo gets moved!"); }
};
struct Bar { Foo foo; };
Bar Meow() { Bar bar; return bar; }
int main() { Bar bar(Meow()); }
On VS11 Beta, in debug mode, this prints:
Foo gets built!
Foo gets copied!
Foo gets copied!
I checked the standard and Bar
seems to meet all requirements to have a default move constructor automatically generated, yet that doesn't seem to happen unless there's another reason why the object cannot be moved. I've seen a lot of move and copy constructor related questions around here but I don't think anyone has had this specific issue.
Any pointers on what's going on here? Is this standard behaviour?
解决方案 Unfortunately, VS11 doesn't provide a default move constructor. See Move Semantics in the Remarks section - to quote:
Unlike the default copy constructor, the compiler does not provide a
default move constructor.
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