C ++中的引用

Question

Once I read in a statement that

The language feature that "sealed the deal" to include references is operator overloading.

Why are references needed to effectively support operator overloading?? Any good explanation?

Answer 1

Here's what Stroustrup said in "The Design and Evolution of C++" (3.7 "references"):

References were introduced primarily to support operator overloading. ...

C passes every function argument by value, and where passing an object by value would be inefficient or inappropriate the user can pass a pointer . This strategy doesn't work where operator overloading is used. In that case, notational convenience is essential because users cannot be expected to insert address-of operators if the objects are large. For example:

  a = b - c; 

is acceptable (that is, conventional) notation, but

  a = &b - &c; 

is not. Anyway, &b - &c already has a meaning in C, and I didn't want to change that.

Answer 2

An obvious example would be the typical overload of ">>" as a stream extraction operator. To work as designed, this has to be able to modify both its left- and right-hand arguments. The right has to be modified, because the primary purpose is to read a new value into that variable. The left has to be modified to do things like indicating the current status of the stream.

In theory, you could pass a pointer as the right-hand argument, but to do the same for the left argument would be problematic (eg when you chain operators together).

Edit: it becomes more problematic for the left side partly because the basic syntax of overloading is that x@y (where "@" stands for any overloaded operator) means x.opertor@(y) . Now, if you change the rules so you somehow turn that x into a pointer, you quickly run into another problem: for a pointer, a lot of those operators already have a valid meaning separate from the overload -- eg, if I translate x+2 as somehow magically working with a pointer to x , then I've produced an expression that already has a meaning completely separate from the overload. To work around that, you could (for example) decide that for this purpose, you'd produce a special kind of pointer that didn't support pointer arithmetic. Then you'd have to deal with x=y -- so the special pointer becomes one that you can't modify directly either, and any attempt at assigning to it ends up assigning to whatever it points at instead.

We've only restricted them enough to support two operator overloads, but our "special pointer" is already about 90% of the way to being a reference with a different name...

Answer 3

References constitute a standard means of specifying that the compiler should handle the addresses of objects as though they were objects themselves. This is well-suited to operator overloading because operations usually need to be chained in expressions; to do this with a uniform interface, ie, entirely by reference, you often would need to take the address of a temporary variable, which is illegal in C++ because pointers make no guarantee about the lifetimes of their referents. References, on the other hand, do. Using references tells the compiler to work a certain kind of very specific magic (with const references at least) that preserves the lifetime of the referent.

Answer 4

Typically when you're implementing an operator you want to operate directly on the operand -- not a copy of it -- but passing a pointer risks that you could delete the memory inside the operator. (Yes, it would be stupid, but it would be a significant danger nonetheless.) References allow for a convenient way of allowing pointer-like access without the "assignment of responsibility" that passing pointers incurs.