My C++ teacher thinks that the * operator in standard C++ is "already overloaded," because it can mean indirection or multiplication depending on the context. He got this from C++ Primer Plus, which states:
Actually, many C++ (and C) operators already are overloaded. For example, the * operator, when applied to an address, yields the value stored at that address. But applying * to two numbers yields the product of the values. C++ uses the number and type of operands to decide which action to take. (pg 502, 5th ed)
At least one other textbook says much the same. So far as I can tell, this is not true; unary * is a different operator from binary *, and the mechanism by which the compiler disambiguates them has nothing to do with operator overloading.
Who is right?
Both are right as the question depends on context and the meaning of the word overloading.
"Overloading" can take a common meaning of "same symbol, different meaning" and allow all uses of "*" including indirection and multiplication, and any user-defined behavior.
"Overloading" can be used to apply to C++'s official operator overloading functionality, in which case indirection and multiplication are indeed different.
ADDENDUM: See Steve's comment below, on "operator overoading" versues "token overloading".
I believe you are. The dereference op and the mult. op are different operators, even if written the same. same goes for +,-,++,--, and any other I may have forgotten.
I believe the book, in this paragraph, refers to the word "overloaded" as "used in more than 1 way", but not by the user. So you could also consider the book as being correct... also depends if you're referring to the overloading of the * operator or of the multiplication operator (for example).
It's overloaded in the sense that the same character is used to mean different things in different places (e.g. pointer dereference, multiplication between ints, multiplication with other built-in types, etc.).
Generally, though, "operator overloading" refers to defining an operator (that has the same symbol as a built-in one) using custom code so that it does something interesting with a user defined type.
So... you're both right :-)
Related
This question already has answers here:
Why is it called operator overloading?
(2 answers)
Closed 4 years ago.
In C++, why is operator overloading called "overloading"?
To me, this seems more like "override."
Because you're not changing the meaning of +, -, *, / etc. with respect to the fundamental datatypes. You can't change what those means for char, short, int, float, etc. Therefore, you're not truly overriding anything.
You are instead expanding the meaning of them to new contexts, which seems to fit with the term "overloading": you've loaded the symbols onto new meanings they did not previously have.
This is pretty subjective, and not easy to answer in specific terms.
But generally we use "override" to mean "replacing the behaviour of a function with another behaviour", as in when you have a polymorphic class hierarchy, and a call to a function whose various implementations are virtual can result in totally different behaviour. Certainly that's what the standard means by the term.
Isn't this also what happens with overloading? Kind of. But usually when you overload a function, you do so to give it different parameter lists, but would still expect each implementation to perform the same job. It doesn't have to, but one expects it to.
Similarly with overloaded operators, if you're overloading say operator+ then generally we expect that it actually still just does the normal, conventional "addition" logic — but overloaded so that it can take an argument of your new class type, instead of the existing overloads that take built-in types.
In practice, that breaks down a bit, because even the standard library makes operator<< mean something completely different (among other examples).
Still, the actual task of creating these new operators is accomplished by what the language considers to be function overloading (particularly as no virtual calls are involved at all).
In short, you're arguably not entirely wrong, but it's pretty arbitrary and this is what we ended up with.
I am reading through the following document,
https://code.google.com/p/go-wiki/wiki/GoForCPPProgrammers
and found the statement below a bit ambiguous:
Unlike in C++, new is a function, not an operator; new int is a syntax error.
In C++ we implement operators as functions, e.g. + using operator+.
So what is the exact difference of operator vs function in programming languages in general?
The actual distinction between functions and operators depends on the programming language. In plain C, operators are a part of the language itself. One cannot add an operator, nor change the behavior of an existing operator. This is not the case with C++, where operators are resolved to functions.
From a totally different point of view, consider Haskell, where ANY (binary) function may be treated as a binary operator:
If you don't speak Haskell, but know about dot products, this example should still be fairly straight-forward. Given:
dotP :: (Double, Double) -> (Double, Double) -> Double
dotP (x1, y1) (x2, y2) = x1 * x2 + y1 * y2
Both
dotP (1,2) (3,4)
and
(1,2) `dotP` (3,4)
will give 11.
To address the quote in the Go documentation: The Go developers are simply stressing that where in C++, one would treat new as a keyword with its own syntax, one should treat new in Go as any other function.
Although I still think the question is basically a duplicate of Difference between operator and function in C++?, it may be worthwhile to clarify what the difference means in the specific context you quoted.
The point there is that a function in C++ is something that has a name and possibly function arguments, and is called using this syntax:
func(arg1,arg2,...)
In other words, the name first, then a round bracket, then the comma-separated list of arguments. This is the function call syntax of C++.
Whereas an operator is used in the way described by clause 5 of the Standard. The details of the syntax vary depending on the kind of operator, e.g. there are unary operators like &, binary operators like +, * etc.; there is also the ternary conditional operator ? :, and then there are special keywords like new, delete, sizeof, some of which translate to function calls for user-defined types, but they don't use the function call syntax described above. I.e. you don't call
new(arg1,arg2,...)
but instead, you use a special "unary expression syntax" (§5.3), which implies, among other things, that there are no round brackets immediately after the keyword new (at least, not necessarily).
It's this syntactic difference that the authors talk about in the section you quoted.
"What is the difference between operators and functions?"
Syntax. But in fact, it's purely a convention with regards to
the language: in C++, + is an infix operator (and only
operators can be infix), and func() would be a function. But
even this isn't always true: MyClass::operator+() is
a function, but it can, and usually is invoked using the
operator syntax.
Other languages have different rules: in languages like Lisp,
there is no real difference. One can distinguish between
built-in functions vs. user defined functions, but the
distinction is somewhat artificial, since you could easily
extend lisp to add additional built-in functions. And there are
languages which allow using the infix notation for user defined
functions. And languages like Python map between them: lhs
+ rhs maps to the function call lhs.__add__( rhs ) (so
"operators" are really just syntactic sugar).
I sum, there is not rule for programming languages in general.
There are simply two different words, and each language is
free to use them as it pleases, to best describe the language.
So what is the exact difference of operator vs function in programming languages in general?
It is broad. In an abstract syntax tree, operators are unary, binary or sometimes ternary nodes - binding expressions together with a certain precedence e.g. + has lower precedence than *, which in turn has lower precedence than new.
Functions are a much more abstract concept. As a primitive, they are typed subroutine entrypoints that depending on the language can be used as rvalues with lexical scope.
C++ allows to override (overload) operators with methods by means of dynamically dispatching operator evaluation to said methods. This is a language "feature" that - as the existence of this question implies - mostly confuses people and is not available in Go.
operators are part of c++ language syntax, in C++ you may 'overload' them as functions if you dont want the default behaviour, For complex types or user defined types , Language may not have the semantic of the operator known , So yuser can overload them with thier own implementation.
I'm designing my own programming language (called Lima, if you care its on www.btetrud.com), and I'm trying to wrap my head around how to implement operator overloading. I'm deciding to bind operators on specific objects (its a prototype based language). (Its also a dynamic language, where 'var' is like 'var' in javascript - a variable that can hold any type of value).
For example, this would be an object with a redefined + operator:
x =
{ int member
operator +
self int[b]:
ret b+self
int[a] self:
ret member+a
}
I hope its fairly obvious what that does. The operator is defined when x is both the right and left operand (using self to denote this).
The problem is what to do when you have two objects that define an operator in an open-ended way like this. For example, what do you do in this scenario:
A =
{ int x
operator +
self var[b]:
ret x+b
}
B =
{ int x
operator +
var[a] self:
ret x+a
}
a+b ;; is a's or b's + operator used?
So an easy answer to this question is "well duh, don't make ambiguous definitions", but its not that simple. What if you include a module that has an A type of object, and then defined a B type of object.
How do you create a language that guards against other objects hijacking what you want to do with your operators?
C++ has operator overloading defined as "members" of classes. How does C++ deal with ambiguity like this?
Most languages will give precedence to the class on the left. C++, I believe, doesn't let you overload operators on the right-hand side at all. When you define operator+, you are defining addition for when this type is on the left, for anything on the right.
In fact, it would not make sense if you allowed your operator + to work for when the type is on the right-hand side. It works for +, but consider -. If type A defines operator - in a certain way, and I do int x - A y, I don't want A's operator - to be called, because it will compute the subtraction in reverse!
In Python, which has more extensive operator overloading rules, there is a separate method for the reverse direction. For example, there is a __sub__ method which overloads the - operator when this type is on the left, and a __rsub__ which overloads the - operator when this type is on the right. This is similar to the capability, in your language, to allow the "self" to appear on the left or on the right, but it introduces ambiguity.
Python gives precedence to the thing on the left -- this works better in a dynamic language. If Python encounters x - y, it first calls x.__sub__(y) to see if x knows how to subtract y. This can either produce a result, or return a special value NotImplemented. If Python finds that NotImplemented was returned, it then tries the other way. It calls y.__rsub__(x), which would have been programmed knowing that y was on the right hand side. If that also returns NotImplemented, then a TypeError is raised, because the types were incompatible for that operation.
I think this is the ideal operator overloading strategy for dynamic languages.
Edit: To give a bit of a summary, you have an ambiguous situation, so you really only three choices:
Give precedence to one side or the other (usually the one on the left). This prevents a class with a right-side overload from hijacking a class with a left-side overload, but not the other way around. (This works best in dynamic languages, as the methods can decide whether they can handle it, and dynamically defer to the other one.)
Make it an error (as #dave is suggesting in his answer). If there is ever more than one viable choice, it is a compiler error. (This works best in static languages, where you can catch this thing in advance.)
Only allow the left-most class to define operator overloads, as in C++. (Then your class B would be illegal.)
The only other option is to introduce a complex system of precedence to the operator overloads, but then you said you want to reduce the cognitive overhead.
I'm going to answer this question by saying "duh, don't make ambiguous definitions".
If I recreate your example in C++ (using a function f instead of the + operator and int/float instead of A/B, but there really isn't much difference)...
template<class t>
void f(int a, t b)
{
std::cout << "me! me! me!";
}
template<class t>
void f(t a, float b)
{
std::cout << "no, me!";
}
int main(void)
{
f(1, 1.0f);
return 0;
}
...the compiler will tell me precisely that: error C2668: 'f' : ambiguous call to overloaded function
If you create a language powerful enough, it's always going to be possible to create things in it that don't make sense. When this happens, it's probably ok to just throw up your hands and say "this doesn't make sense".
In C++, a op b means a.op(b), so it's unambigious; the order settles it. If, in C++, you want to define an operator whose left operand is a built-in type, then the operator has to be a global function with two arguments, not a member; again, though, the order of the operands determines which method to call. It is illegal to define an operator where both operands are of built-in types.
I would suggest that given X + Y, the compiler should look for both X.op_plus(Y) and Y.op_added_to(X); each implementation should include an attribute indicating whether it should be a 'preferred', 'normal', 'fallback' implementation, and optionally also indicating that it is "common". If both implementations are defined, and they implementations are of different priorities (e.g. "preferred" and "normal"), use the type to select a preference. If both are defined to be of the same priority, and both are "common", favor the X.op_plus(Y) form. If both are defined with the same priority and they are not both "common", flag an error.
I would suggest that the ability to prioritize overloads and conversions would IMHO a very important feature for a language to have. It is not helpful for languages to squawk about ambiguous overloads in cases where both candidates would do the same thing, but languages should squawk in cases where two possible overloads would have different meanings, each of which would be useful in certain contexts. For example, given someFloat==someDouble or someDouble==someLong, a compiler should squawk, since there can be usefulness to knowing whether the numerical quantities represented by two values match, and there can also be usefulness in knowing whether the left-hand operand holds the best possible representation (for its type) of the value in the right-hand operand. Java and C# do not flag ambiguity in either case, opting instead to use the first meaning for the first expression and the second for the second, even though either meaning might be useful in either case. I would suggest that it would be better to reject such comparisons than to have them implement inconsistent semantics.
Overall, I'd suggest as a philosophy that a good language design should let a programmer indicate what's important and what isn't. If a programmer knows that certain "ambiguities" aren't problems, but other ones are, it should be easy to have the compiler flag the latter but not the former.
Addendum
I looked briefly through your proposal; it sees you're expecting bindings to be fully dynamic. I've worked with a language like that (HyperTalk, circa 1988) and it was "interesting". Consider, for example, that "2X" < "3" < 4 < 10 < "11" < "2X". Double dispatch can sometimes be useful, but only in cases where operators overloads with different semantics (e.g. string and numeric comparisons) are limited to operating on disjoint sets of things. Forbidding ambiguous operations at compile time is a good thing, since the programmer will be in a position to specify what's intended. Having such ambiguity trigger a run-time error is a bad thing, because the programmer may be long gone by the time an error surfaces. Consequently, I really can't offer any advice for how to do run-time double dispatch for operators except to say "don't", unless at compile time you restrict the operands to combinations where any possible overload would always have the same semantics.
For example, if you had an abstract "immutable list of numbers" type, with a member to report the length or return the number at a particular index, you could specify that two instances are equal if they have the same length, and every for every index they return the same number. While it would be possible to compare any two instances for equality by examining every item, that could be inefficient if e.g. one instance was a "BunchOfZeroes" type which simply held an integer N=1000000 and didn't actually store any items, and the other was an "NCopiesOfArray" which held N=500000 and {0,0} as the array to be copied. If many instances of those types are going to be compared, efficiency could be improved by having such comparisons invoke a method which, after checking overall array length, checks whether the "template" array contains any non-zero elements. If it doesn't, then it can be reported as equal the bunch-of-zeroes array without having to perform 1,000,000 element comparisons. Note that the invocation of such a method by double dispatch would not alter the program's behavior--it would merely allow it to execute more quickly.
It is reasonable that sizeof and typeid cannot be overloaded, but I can't see the harm in overloading ?:, .* and .. Are there technical reasons for this?
To quote Bjarne Stroustrup:
There is no fundamental reason to
disallow overloading of ?:. I just
didn't see the need to introduce the
special case of overloading a ternary
operator. Note that a function
overloading expr1?expr2:expr3 would
not be able to guarantee that only one
of expr2 and expr3 was executed.
...
Operator . (dot) could in principle be
overloaded using the same technique as
used for ->. However, doing so can
lead to questions about whether an
operation is meant for the object
overloading . or an object referred to
by . ... This problem can be solved in
several ways. At the time of
standardization, it was not obvious
which way would be best.
Source
If you overload ., how would you access class members? What would be the meaning of obj.data?
What would the syntax be?
In fact, there are good reasons for not overloading any operator
which doesn't evaluate all of its operands: you shouldn't
overload && or || either (except in special cases). You can't
simulate this with an overloaded operator. Consider something
like:
p != NULL ? defaultValue : p->getValue()
where the type of defaultValue or p->getValue() causes overload
resolution to pick up your overload. It's a common idiom, but
it can't be made to work if you overloaded ?:.
Here's some reading material C++ FAQ Lite :)
In general there would be no benefit to overloading the operators above. What additional semantics would you be trying to implement?
The reason for overloading operators is to provide intuitive syntax to the user of your class. For example, it makes sense to overload + and += for strings. It's obvious to another developer what that means.
It's really not obvious what you would overload ?: for ... That said there are no technical reasons I am aware of that prevented these operators from being overloaded.
Overloading the -> operator allows for reference counted pointers to be created such as boost::shared_ptr. The concept of 'negating' an object might have different meanings in different contexts, so it's reasonable to occasionally overload this operator.
Defining "operator bool" is enough for ?: to work.
For operator . think of this: SomeClass."SomeString!!"
These overloadings prohibit compiler's lexer from parsing the file correctly.
The reason you can overload most operators is to be able to simulate built in types. Since none of the built in types can use the . operator, it wouldn't serve any purpose. operator* and operator-> are there so you can make your own pointer classes. All the math and boolean operators are there to be able to make your own numeric classes.
The C++ standard defines the expression using subscripts as a postfix expression. AFAIK, this operator always takes two arguments (the first is the pointer to T and the other is the enum or integral type). Hence it should qualify as a binary operator.
However MSDN and IBM does not list it as a binary operator.
So the question is, what is subscript operator? Is it unary or binary? For sure, it is not unary as it is not mentioned in $5.3 (at least straigt away).
What does it mean when the Standard mentions it's usage in the context of postfix expression?
I'd tend to agree with you in that operator[] is a binary operator in the strictest sense, since it does take two arguments: a (possibly implicit) reference to an object, and a value of some other type (not necessarily enumerated or integral). However, since it is a bracketing operator, you might say that the sequence of tokens [x], where x might be any valid subscript-expression, qualifies as a postfix unary operator in an abstract sense; think currying.
Also, you cannot overload a global operator[](const C&, size_t), for example. The compiler complains that operator[] must be a nonstatic member function.
You are correct that operator[] is a binary operator but it is special in that it must also be a member function.
Similar to operator()
You can read up on postfix expressions here
I just found an interesting article about operator[] and postfix expression, here
I think it's the context that [] is used in that counts. Section 5.2.1 the symbol [] is used in the context of a postfix expression that is 'is identical (by definition) to *((E1)+(E2))'. In this context, [] isn't an operator. In section 13.5.5 its used to mean the subscripting operator. In this case it's an operator that takes one argument. For example, if I wrote:
x = a[2];
It's not necessarily the case that the above statement evaluates to:
x = *(a + 2);
because 'a' might be an object. If a is an object type then in this context, [] is used as an subscript operator.
Anyway that's the best explanation I can derive from the standard that resolves apparent contradictions.
If you take a close look to http://en.wikipedia.org/wiki/Operators_in_C_and_C%2B%2B it will explain you that standard C++ recognize operator[] to be a binary operator, as you said.
Operator[] is, generally speaking, binary, and, despite there is the possibility to make it unary, it should always be used as binary inside a class, even because it has no sense outside a class.
It is well explained in the link I provided you...
Notice that sometimes many programmers overload operators without think too much about what they are doing, sometimes overloading them in an incorrect manner; the compiler is ease is this and accept it, but, probably, it was not the correct way to overload that operator.
Following guides like the one I provided you, is a good way to do things in the correct manner.
So, always beware examples where operators are overloaded without a good practice (out of standard), refer, first to the standard methods, and use those examples that are compliant to them.