C++ 非 const 左值引用
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Non const lvalue references
提问by Mars
Why can you do this
为什么你可以这样做
int a;
const double &m = a;
But when you do this
但是当你这样做时
int a;
double &m = a;
you get an error?
你有错误吗？
error: non-const lvalue reference to type 'double' cannot bind to a value of unrelated type 'int'
Edit:
编辑：
To be more specific I am trying to understand the reason non-const references can't bind temp objects.
更具体地说，我试图了解非常量引用无法绑定临时对象的原因。
采纳答案by Mahesh
That is because a temporary can not bind to a non-const reference.
那是因为临时不能绑定到非常量引用。
double &m = a;
ais of type intand is being converted to double. So a temporary is created. Same is the case for user-defined types as well.
a是 类型int并且正在转换为double. 所以临时创建了。用户定义的类型也是如此。
Foo &obj = Foo(); // You will see the same error message.
But in Visual Studio, it works fine because of a compiler extension enabled by default. But GCC will complain.
但在 Visual Studio 中，由于默认启用编译器扩展，它运行良好。但是海湾合作委员会会抱怨。
回答by xinnjie
Because making modification on a temporary is meaningless, c++ don't want you to bind non-const reference to a temporary. For example.
因为对临时对象进行修改是没有意义的，所以 C++ 不希望您将非常量引用绑定到临时对象。例如。
int a;
double &m = a; // caution:this does not work.
What if it works?
ais of type int and is being converted to double. So a temporary is created.
如果有效呢？
a是 int 类型，正在转换为 double。所以临时创建了。
You can modify m, which is bound to a temporary, but almost nothing happens.After the modification, variable adoes not change(what's worse, you might think ahas changed, which may cause problems).
你可以修改m，它绑定了一个临时的，但几乎没有任何反应。修改后，变量a没有改变（更糟糕的是，你可能认为a已经改变了，这可能会导致问题）。
