templated operator "+" overloading and enum initialization [constexpr][msvc] - c++

Please, consider the following code:
template<typename T>
T operator+(const T& lvh, const T& rvh)
{
return lvh;
}
struct enum_container {
enum {
item1 = 1,
item2 = 2,
item3 = item1 + item2
};
};
int main() {
return 0;
}
Compiling with MS VS 2022 gives the following error:
error C2131: expression did not evaluate to a constant
message : failure was caused by call of undefined function or one not declared 'constexpr'
message : see usage of 'operator +'
The problem applies to the third element in enum.
Why cl tries to substitute template for the constant expression?
Compilation with gcc gives no errors.
I can eliminate the error with cl adding type specifier to the enum, like this:
enum : int {
...
}
But I can`t do this in the project I work with. Both enum and overload declared by 3rd party headers. All I can do is to change includ order. But I'm still wondering why it works that way...

To me it looks like MSVC is correct here.
First, because template argument deduction should happen before overload resolution to form a set of candidate functions:
Before overload resolution begins, the functions selected by name
lookup and template argument deduction are combined to form the set of
candidate functions
And then, since the non-member candidates (in this case template specialisation of the operator+ overload with the enum_container::<anonymous_enum> parameters) precede built-in candidates (conversion to the underlying built-in type, for which a user-defined overload would not be possible), the compiler is meant to resolve the overload to the instance of the template with the user-defined enum-type:
non-member candidates: For the operators where operator overloading
permits non-member forms, all declarations found by unqualified name
lookup of operator# in the context of the expression (which may
involve ADL), except that member function declarations are ignored and
do not prevent the lookup from continuing into the next enclosing
scope. If both operands of a binary operator or the only operand of a
unary operator has enumeration type, the only functions from the
lookup set that become non-member candidates are the ones whose
parameter has that enumeration type (or reference to that enumeration
type)
built-in candidates: For operator,, the unary operator&, and the operator->, the set of built-in candidates is empty. For other
operators built-in candidates are the ones listed in built-in operator
pages as long as all operands can be implicitly converted to their
parameters. If any built-in candidate has the same parameter list as a
non-member candidate that isn't a function template specialization, it
is not added to the list of built-in candidates. When the built-in
assignment operators are considered, the conversions from their
left-hand arguments are restricted: user-defined conversions are not
considered.
Thus, the fact that GCC/Clang ignore the existence of the user-defined overload of operator+ under the enum definition doesn't seem to be compliant with the rules listed above. The same expression inside of a block scope causes exactly same error for all compilers (and I couldn't find any rationale why this might be different for the scope of enum definition):
struct enum_container {
enum {
item1 = 1,
item2 = 2,
item3 = item1 + item2
};
};
template<typename T>
T operator+(const T& lhs, const T&) {
return lhs;
}
int main() {
// error: call to non-'constexpr' function 'T operator+(const T&, const T&)
constexpr auto enumval = enum_container::item1 + enum_container::item2;
return 0;
}

Related

Casting to void to avoid use of overloaded user-defined Comma operator

I am learning about templates in C++, and came across an example where casting to void is used:
template<typename T>
auto func (T const& t) -> decltype( (void)(t.size()), T::size_type() )
{
return t.size();
}
In the explanation it is written that:
The cast of the expression to void is to avoid the possibility of a user-defined comma operator overloaded for the type of the expressions.
My question(s) is/are:
How can a cast to void be used to "avoid the possibility of a user-defined comma operator overloaded for the type of the expressions"? I mean, can anyone give any example where if we don't use void then this code would give an error? For example, let's say we have a class called SomeClass which has overloaded the comma operator. Now, can this become a problem if we don't use void?
Can static_cast be used in this case instead of a C style cast? For example, something like static_cast<void>(t.size()). I am reading examples that use C++17 features, and so I wonder why the author has used a C style cast in this case.
I have read What does casting to `void` really do?, from which I get the impression that if we use (void)x then this means to suppress compiler warnings, and also means "ignore the value of x". But then I can't understand the difference between the expression x and (void)x.
Consider this pathological type:
struct foo {
struct size_type {
bool operator,(size_type) { return false;}
};
size_type size() { return {};}
};
It does have a size_type and it does have a size() method. However, without the cast to void, the template does not deduce the right return type, because decltype( (t.size()), typename T::size_type() ) is bool :
#include <type_traits>
template<typename T>
auto func (T const& t) -> decltype( (t.size()), typename T::size_type() )
{
return t.size();
}
struct foo {
struct size_type {
bool operator,(size_type) { return false;}
};
size_type size() const { return {};}
};
int main()
{
func(foo{});
}
Results in error:
<source>:6:8: error: no viable conversion from returned value of type 'foo::size_type' to function return type 'decltype((t.size()) , typename foo::size_type())' (aka 'bool')
return t.size();
^~~~~~~~
<source>:20:4: note: in instantiation of function template specialization 'func<foo>' requested here
func(foo{});
^
1 error generated.
ASM generation compiler returned: 1
<source>:6:8: error: no viable conversion from returned value of type 'foo::size_type' to function return type 'decltype((t.size()) , typename foo::size_type())' (aka 'bool')
return t.size();
^~~~~~~~
<source>:20:4: note: in instantiation of function template specialization 'func<foo>' requested here
func(foo{});
^
A static_cast can be used. However, as nothing is actually being cast (it is an unevaluated context), the c-style cast does not do much harm. Note how by using the cast to void the user defined operator, is bypassed and the correct return type is deduced: https://godbolt.org/z/jozx1YGWr. This is because void, whatever uses the built-in operator, whose result is of same type as whatever. You cannot override void, whatever with a user-defined operator,; there is no syntax that works.
I suppose the code is merely to illustrate this one effect, because even with the cast to void one can make up other examples that fail (eg the template does not explicitly test for t.size() actually returning T::size_type).
See also the section on "Rarely overloaded operators" here:
The comma operator, operator,. Unlike the built-in version, the overloads do not sequence their left operand before the right one. (until C++17) Because this operator may be overloaded, generic libraries use expressions such as a,void(),b instead of a,b to sequence execution of expressions of user-defined types. The boost library uses operator, in boost.assign, boost.spirit, and other libraries. The database access library SOCI also overloads operator,.

Overloading operator== with `&&` and `const` qualifier cause ambiguity in C++20

Consider the struct S with two operator== overloads of same && qualifier and different const qualifier:
struct S {
bool operator==(const S&) && {
return true;
}
bool operator==(const S&) const && {
return true;
}
};
If I compare the two S with operator==:
S{} == S{};
gcc and msvc accept this code, clang rejects it with:
<source>:14:7: error: use of overloaded operator '==' is ambiguous (with operand types 'S' and 'S')
S{} == S{};
~~~ ^ ~~~
Why does clang think there is an ambiguous overload resolution here? Shouldn't the non-const one be the best candidate in this case?
Similarly, if I compare two S with the synthesized operator!=:
S{} != S{};
gcc still accept this code, but msvc and clang doesn't:
<source>:14:7: error: use of overloaded operator '!=' is ambiguous (with operand types 'S' and 'S')
S{} != S{};
~~~ ^ ~~~
It seems weird that the synthesized operator!= suddenly causes the ambiguity for msvc. Which compiler is right?
The example would be unambiguous in C++17. C++20 brings change:
[over.match.oper]
For a unary operator # with an operand of type cv1 T1, and for a binary operator # with a left operand of type cv1 T1 and a right operand of type cv2 T2, four sets of candidate functions, designated member candidates, non-member candidates, built-in candidates, and rewritten candidates, are constructed as follows:
...
For the operator ,, the unary operator &, or the operator ->, the built-in candidates set is empty. For all other operators, the built-in candidates include all of the candidate operator functions defined in [over.built] that, compared to the given operator,
have the same operator name, and
accept the same number of operands, and
accept operand types to which the given operand or operands can be converted according to [over.best.ics], and
do not have the same parameter-type-list as any non-member candidate that is not a function template specialization.
The rewritten candidate set is determined as follows:
...
For the equality operators, the rewritten candidates also include a synthesized candidate, with the order of the two parameters reversed, for each non-rewritten candidate for the expression y == x.
Thus, the rewritten candidate set includes these:
implicit object parameter
|||
(S&&, const S&); // 1
(const S&&, const S&); // 2
// candidates that match with reversed arguments
(const S&, S&&); // 1 reversed
(const S&, const S&&); // 2 reversed
The overload 1 is better match than 2, but the synthesised reversed overload of 1 is ambiguous with the original non-reversed overload because both have const conversion to one parameter. Note that this is actually ambiguous even if overload 2 doesn't exist.
Thus, Clang is correct.
This is also covered by the informative compatibility annex:
Affected subclause: [over.match.oper] Change: Equality and inequality expressions can now find reversed and rewritten candidates.
Rationale: Improve consistency of equality with three-way comparison and make it easier to write the full complement of equality
operations.
Effect on original feature: Equality and inequality expressions between two objects of different types, where one is convertible to
the other, could invoke a different operator. Equality and inequality
expressions between two objects of the same type could become
ambiguous.
struct A {
operator int() const;
};
bool operator==(A, int); // #1
// #2 is built-in candidate: bool operator==(int, int);
// #3 is built-in candidate: bool operator!=(int, int);
int check(A x, A y) {
return (x == y) + // ill-formed; previously well-formed
(10 == x) + // calls #1, previously selected #2
(10 != x); // calls #1, previously selected #3
}

C++ "error: use of overloaded operator '*' is ambiguous" with seemingly only one match

I am trying to use custom operators in C++ for a project I am working on. This project uses the ROCm/HIP stack (so, under the hood, the clang compiler).
Here's the error message:
src/zlatrd.cpp:359:32: error: use of overloaded operator '*' is ambiguous (with operand types 'magmaDoubleComplex' (aka 'hip_complex_number<double>') and 'float')
alpha = tau[i] * -0.5f * value;
~~~~~~ ^ ~~~~~
./include/magma_operators.h:190:1: note: candidate function
operator * (const magmaDoubleComplex a, const double s)
^
./include/magma_operators.h:183:1: note: candidate function
operator * (const magmaDoubleComplex a, const magmaDoubleComplex b)
^
./include/magma_operators.h:437:1: note: candidate function
operator * (const magmaFloatComplex a, const float s)
^
./include/magma_operators.h:430:1: note: candidate function
operator * (const magmaFloatComplex a, const magmaFloatComplex b)
^
It seems to me that it is not ambiguous; it should select the third candidate function, as the argument is a float.
Here is the type definition for the hip_complex_number template:
template <typename T>
struct hip_complex_number
{
T x, y;
template <typename U>
hip_complex_number(U a, U b)
: x(a)
, y(b)
{
}
template <typename U>
hip_complex_number(U a)
: x(a)
, y(0)
{
}
hip_complex_number()
: x(0)
, y(0)
{
}
};
I notice it has an implicit constructor that will convert a float, but I assumed that given a candidate function that matches the type exactly (not including the const modifier), that it would obviously select that function overload over those which require an implicit cast.
EDIT: Also, I know that by default C/C++ convert from float/double to each other if the function is defined in that way, so 'matches the type exactly' was definitely not the right wording.
Can someone explain why C++ thinks this is ambiguous?
EDIT: People have asked for the definition of magmaFloatComplex, which is hip_complex_number<float>
Please note that I don't know anything about the library that these types are from. I will explain the ambiguity purely based on the information in the question.
The first and third overload are ambiguous.
In the overload operator * (const magmaDoubleComplex a, const double s) a floating-point promotion from float to double is required in the second argument.
In the overload operator * (const magmaFloatComplex a, const float s) a user-defined conversion to an unrelated type from magmaDoubleComplex to magmaFloatComplex is required. This conversion is possible, because of the non-explicit converting constructor
template <typename U>
hip_complex_number(U a)
The corresponding other parameters don't need any conversion aside from potentially lvalue-to-rvalue conversions or user-defined conversion to the same type, which are considered exact match.
Exact match is better than either user-defined conversion to an unrelated type or floating-point promotion, meaning that each overload has one parameter that is better than the other one's and one that is worse than the other one's.
Overload resolution is ambiguous if not at least one overload has all parameters not worse than all other overload's parameters. Therefore the two mentioned overloads are ambiguous here.
The second overload has exact match in the first argument and requires a user-defined conversion to unrelated type in the second argument, which is again possible because of the converting constructor mentioned above. However the floating-point promotion of the first overload is considered better than a user-defined conversion to unrelated type and therefore the second overload looses against the first one in overload resolution, but would be ambiguous with the third one as well.
The fourth overload is worse than all the others, because it requires user-defined conversions to unrelated types in both parameters.
Note that if overload 3 would be selected as expect in your question, it would result in an error, because the converting constructor chosen for magmaFloatComplex will try to initialize the x member which is of type float with a magmaDoubleComplex, which (at least based on your shown code) doesn't have a conversion operator to float.

Overloading conversion operator template

Consider the following simple example
struct C
{
template <typename T> operator T () {return 0.5;}
operator int () {return 1;}
operator bool () {return false;}
};
int main ()
{
C c;
double x = c;
std::cout << x << std::endl;
}
When compiled with Clang, it gives the following error
test.cpp:11:12: error: conversion from 'C' to 'double' is ambiguous
double x = c;
^ ~
test.cpp:4:5: note: candidate function
operator int () {return 1;}
^
test.cpp:5:5: note: candidate function
operator bool () {return false;}
^
test.cpp:3:27: note: candidate function [with T = double]
template <typename T> operator T () {return 0.5;}
^
1 error generated.
Other compilers generate similar errors, e.g., GCC and Intel iclc
If I remove operator int and operator bool. It compiles fine and work as expected. If only remove one of them, that is keep the template operator and say operator int, then the non-template version is always chosen.
My understanding is that only when the template and non-template overloaded functions are equal in the sense that they are both perfect match or both require the same conversion sequence, the non-template version will be preferred. However in this case, it appears that the compiler does not see the operator template as a perfect match. And when both the bool and int overloading are present, then naturally it considers they are ambiguous.
In summary, my question is that why the operator template is not considered a perfect match in this case?
Let's break this down into two different problems:
1. Why does this generate a compiler error?
struct C
{
operator bool () {return false;}
operator int () {return 1;}
};
As both int and bool can be implicitly converted to double, the compiler can not know which function it should use. There are two functions which it could use and neither one takes precedence over the other one.
2. Why isn't the templated version a perfect match?
struct C
{
template <typename T> operator T () {return 0.5;}
operator int () {return 1;}
};
Why is operator int() called when requesting a double?
The non-template function is called because a non-template function takes precedence in overload resolution. (Overloading function templates)
EDIT:
I was wrong! As Yan Zhou mentioned in his comment, and as it is stated in the link I provided, a perfect match in the templated function takes precedence over the non-templated function.
I tested your code (compiled with g++ 4.7.2), and it worked as expected: it returned 0.5, in other words, the templated function was used!
EDIT2:
I now tried with clang and I can reproduce the behaviour you described. As it works correctly in gcc, this seems to be a bug in clang.
This is interesting. There are two ways to read a critical part of section 13.3.3. The original example should definitely call the function template, but the version where one of the non-templates is removed might be argued to be ambiguous.
13.3.3:
A viable function F1 is defined to be a better function than another viable function F2 if for all arguments i, ICS_i(F1) is not a worse conversion sequence than ICS_i(F2), and then
for some argument j, ICS_j(F1) is a better conversion sequence than ICS_j(F2), or, if not that,
the context is an initialization by user-defined conversion (see 8.5, 13.3.1.5, and 13.3.1.6) and the standard conversion sequence from the return type of F1 to the destination type (i.e., the type of the entity being initialized) is a better conversion sequence than the standard conversion sequence from the return type of F2 to the destination type, or, if not that,
F1 is a non-template function and F2 is a function template specialization, or, if not that,
F1 and F2 are function template specializations, and the function template for F1 is more specialized than the template for F2 according to the partial ordering rules described in 14.5.6.2.
If there is exactly one viable function that is a better function than all other viable functions, then it is the one selected by overload resolution; otherwise the call is ill-formed.
In the example, clang correctly identifies the set of three viable candidate functions:
C::operator int()
C::operator bool()
C::operator double<double>()
The third is a function template specialization. (I don't think the syntax above is legal, but you get the idea: at this point of overload resolution it's not treated as a template, but as a specialization with a definite function type.)
The only Implicit Conversion Sequence on arguments here (ICS1) is the exact match "lvalue C" to "C&" on the implicit parameter, so that won't make a difference.
This example is exactly the situation described in the second bullet, so the function returning double is clearly better than the other two.
Here's where it gets weird: By a very literal reading, operator int is also better than the template specialization, because of the third bullet. "Wait a minute, shouldn't 'better than' be antisymmetric? How can you say F1 is better than F2 AND F2 is better than F1?" Unfortunately, the Standard doesn't explicitly say anything of the sort. "Doesn't the second bullet take priority over the third bullet because of the 'if not that' phrase?" Yes, for constant F1 and F2. But the Standard doesn't say that satisfying the second bullet for (F1,F2) makes the third bullet for (F2,F1) not applicable.
Of course, since operator int is not better than operator bool and vice versa, there is still "exactly one viable function that is a better function than all other viable functions".
I'm not exactly endorsing this weird reading, except maybe to report it as a Standard defect. Going with that would have bizarre consequences (like removing an overload which was not the best from this example changes a program from well-formed to ambiguous!). I think the intent is for the second bullet to be considered both ways before the third bullet is considered at all.
Which would mean the function template should be selected by overload resolution, and this is a clang bug.

misunderstanding the global operator overload rule

I found the URL below that says that
If an operator can be used as either a unary or a binary
operator (&, *, +, and -), you can overload each use separately.
I am working with g++ in Linux and I tried the following and it didn't compile.
int operator+ (const int a,const int b){
std::cout << "MINE"<<std::endl;
return 0;
}
int main(){
char c='c';
std::cout << c+2 << std::endl;
}
The error says
error: ‘int operator+(int, int)’ must have an argument
of class or enumerated type
I was willing to play and see in action the Integer Promotion Rules.
Am I doing something wrong or that URL is valid only for MS or I misunderstood the promotion rule?
The error message indirectly tells you what you need to know -- you are not permitted to overload operators (binary or unary) that act only on built-in types.
For a user-defined type T, you can separately overload binary + (for example by T operator+(T lhs, T rhs)) and unary + (for example by T operator+(T t)). You could also define operator+(T lhs, int rhs), but you can't overload addition of two integers.
Example is incorrect since
n3337 13.5/6
An operator function shall either be a non-static member function or be a non-member function and have
at least one parameter whose type is a class, a reference to a class, an enumeration, or a reference to an
enumeration.