This simple code
bool foo(std::istringstream&stream, std::string&single, char del)
{ return std::getline(stream,single,del); }
compiles with gcc (4.8.2) but not with clang (3.4, using libc++), which complains that there is no viable conversion from std::basic_istream<char, std::char_traits<char> > to bool. However, when I wrap argument to the return statement in a static_cast<bool>(), clang is happy.
This confused me made me wonder whether the above code is well formed or not, i.e. whether gcc or clang is correct. According to cpprefernce std::getline returns a std::basic_istream<char, std::char_traits<char> >, which is inherited from a std::basic_ios which has the type conversion operator bool (since C++11, before it was a type conversion to void*). Shouldn't this conversion operator get selected automatically? (for some reason, I'm more ready to accept that gcc is wrong than clang).
Edit I just figured out that apparently libc++ of llvm declares the conversion operator in question explicit, deeming it invalid for the implicit conversion. Is this in line with the standard?
clang is correct. Since C++11, the conversion from std::basic_ios to bool is indeed required to be explicit.
C++11, [ios.overview]:
explicit operator bool() const;
As the initialization that occurs in function return is the copy-initialization then the explicit conversion operator bool may not be applied implicitly.
From the C++ Standard
2 A conversion function may be explicit (7.1.2), in which case it is
only considered as a user-defined conversion for direct-initialization
(8.5).
So GCC has a bug.
The operator may be applied implicitly when the direct-initialization is used or in special context as the context of the if condition.
Related
I considered a type conversion from one type to the other, which are defined by two way, i.e., type conversion constructor and type conversion function.
struct to_type;
struct from_type{
operator to_type()const;
};
struct to_type{
to_type() = default;
to_type(const from_type&){}
};
from_type::operator to_type()const{return to_type();}
int main(){
from_type From;
to_type To;
To = From;
return 0;
}
gcc (v13.0.0, but seems to be same even in v4.9.4) don't throw any error and just call type conversion constructor in the above code.
On the other hand, clang (v16.0.0, but seems to be same even in v7.6.0) throw a "ambiguous" compile error like the following.
prog.cc:14:10: error: reference initialization of type 'to_type &&' with initializer of type 'from_type' is ambiguous
To = From;
^~~~
prog.cc:3:4: note: candidate function
operator to_type()const;
^
prog.cc:7:4: note: candidate constructor
to_type(const from_type&){}
^
prog.cc:5:8: note: passing argument to parameter here
struct to_type{
^
1 error generated.
It seems to be so curious that two major compiler show different result for this simple code. Is either compiler don't match with the standard for C++? I guess this overload resolution related to [over.ics.rank], but I could not concluded which compiler's behavior match to the standard.
Or do my source code contains undefined behavior?
[ADD 2022-12-21T00:20Z]
Following the comment by Artyer, I tried -pedantic compile option for gcc, and now gcc also output error message.
prog.cc: In function 'int main()':
prog.cc:14:10: error: conversion from 'from_type' to 'to_type' is ambiguous
14 | To = From;
| ^~~~
prog.cc:9:1: note: candidate: 'from_type::operator to_type() const'
9 | from_type::operator to_type()const{return to_type();}
| ^~~~~~~~~
prog.cc:7:4: note: candidate: 'to_type::to_type(const from_type&)'
7 | to_type(const from_type&){}
| ^~~~~~~
prog.cc:5:8: note: initializing argument 1 of 'constexpr to_type& to_type::operator=(to_type&&)'
5 | struct to_type{
| ^~~~~~~
This suggests that at least the default behavior of gcc without -pedantic don't match with the requirements of C++ standard for this source code.
In this case clang is right, this behaviour is defined in [over.best.ics.general], and standard even mentions that such a conversion is ambiguous explicitly under the sample code to [over.best.ics.general]/10 (the scenario under the link is in fact considers another kind of ambiguity, but resolution to user-defined pair of conversion constructor and conversion operator is one of the candidates, so I removed the part of the code with another candidate):
class B;
class A { A (B&);};
class B { operator A (); };
...
void f(A) { }
...
B b;
f(b); // ... an (ambiguous) conversion b → A (via constructor or conversion function)
In order to break the name resolution down, I'd like to represent the conversion sequence (To = From;) as an assignment operator function:
to_type& operator=(to_type&& param) { } // defined implicitly by the compiler
The actual conversion happens when param needs to get into existence out of From argument of type from_type. The compiler then needs to decide step by step:
Which type the conversion sequence from from_type to to_type&& is of? (standard/user defined/ellipsis):
The types from_type is user defined, but to_type&& is of reference type, and reference binding could be considered identity (i.e. standard) conversion. However it's not the case, since from_type and to_type are not the same and cannot be bound directly ([over.ics.ref]/2):
When a parameter of reference type is not bound directly to an argument expression, the conversion sequence is the one required to convert the argument expression to the referenced type according to [over.best.ics].
Without reference binding there is no any other standard conversion sequence that may suit here. Let's consider user-defined conversion. [over.ics.user] gives us the following definition:
A user-defined conversion sequence consists of an initial standard conversion sequence followed by a user-defined conversion ([class.conv]) followed by a second standard conversion sequence. If the user-defined conversion is specified by a constructor ([class.conv.ctor]), the initial standard conversion sequence converts the source type to the type of the first parameter of that constructor. If the user-defined conversion is specified by a conversion function, the initial standard conversion sequence converts the source type to the type of the implicit object parameter of that conversion function.
This sounds about right to me: we need to convert from_type argument to to_type temporary in order for to_type&& to bind to the argument, thus the sequence is either
from_type -> const from_type& for the converting constructor argument to_type(const from_type&) -> to_type&& for the move-assignment operator of to_type& operator=(to_type&&)
OR
from_type -> implicit-object-parameter of type const from_type& for conversion function operator to_type() const -> to_type&& for the move-assignment operator of to_type& operator=(to_type&&).
Now we have two possible conversion sequences of the same kind (user-defined). For this scenario [over.best.ics.general]/10 says the following:
If there are multiple well-formed implicit conversion sequences converting the argument to the parameter type, the implicit conversion sequence associated with the parameter is defined to be the unique conversion sequence designated the ambiguous conversion sequence. For the purpose of ranking implicit conversion sequences as described in [over.ics.rank].
The Ranking implicit conversion sequences documentation then gives the following clues about deciding on which conversion (of the same sequence type) should take precedence for user-defined sequences ([over.ics.rank]/3.3, emphasis mine):
User-defined conversion sequence U1 is a better conversion sequence than another user-defined conversion sequence U2 if they contain the same user-defined conversion function or constructor or they initialize the same class in an aggregate initialization and in either case the second standard conversion sequence of U1 is better than the second standard conversion sequence of U2
Here we go, for both scenarios (with the converting constructor and the conversion function) the second standard conversion is of the same type (a temporary of type to_type to to_type&&), thus the operations are indistinguishable.
Clang is wrong in rejecting the program because an overload with T&& is better match than an overload with const int&. Note that gcc is also wrong because it uses the copy constructor instead of operator to_type() const. See demo. Only msvc is right and both gcc and clang are wrong. MSVC correctly uses the conversion function.
S1 S2
int int&& indistinguishable
int const int& indistinguishable
int&& const int& S1 better
Consider the contrived example:
#include <iostream>
struct to_type;
struct from_type{
operator to_type()const;
};
struct to_type{
to_type() = default;
to_type(const from_type&){std::cout <<"copy ctor";}
};
from_type::operator to_type()const{
std::cout<<"to_type operator";
return to_type();}
void f(to_type&&){}
int main(){
from_type From;
f(From); //valid and this should use from_type::operator to_type() const
}
Demo
The above program is rejected by clang(with the same error as you're getting) but accepted by gcc and msvc. Note that even though gcc accepts the above program it is still wrong because the conversion function should be used instead of the copy ctor. MSVC on the other hand correctly uses the conversion function.
Consider the code below:
#include <cstdio>
struct A
{
A(){}
explicit A(const A&) {std::puts("copy");}
};
int main()
{
A a;
true ? a : A();
return 0;
}
As I understand the ternary would try to copy a and should fail since the copy constructor is explicit, however gcc compiles this just fine and creates a copy. Clang spits out an error as expected.
Is this a bug in gcc?
I'm using gcc 8.1 and clang 7.0, in c++17 mode, but I also tried all versions of gcc in compiler explorer in c++98 mode, and they all behave the same.
Clang is right to reject it, and it is indeed a GCC bug. I'll quote n4659 (closest document I have to to the C++17 standard) for simplicity.
First and foremost, the type of of the conditional expression in your example, as specified by [expr.cond] ¶6 must be a prvalue of type A.
Now, according to [expr.cond] ¶7, emphasis mine:
Lvalue-to-rvalue, array-to-pointer, and function-to-pointer standard
conversions are performed on the second and third operands.
a must be able to undergo the lvalue-to-rvalue conversion. Which for a is specified in [conv.lval] ¶3.2 (again, emphasis mine) as
Otherwise, if T has a class type, the conversion copy-initializes the
result object from the glvalue.
Copy initialization of an A from an A, in any context, should pick a converting constructor in overload resolution ([over.match.copy] ¶1.1):
The converting constructors of T are candidate functions.
And an explicit copy constructor is not a converting constructor ([class.conv.ctor] ¶3)
A non-explicit copy/move constructor ([class.copy]) is a converting constructor.
A conforming C++ implementation cannot accept the conditional expression your wrote as well-formed.
The following code compiles fine on all compilers I've checked (clang, mingw, g++) other than MSVC.
enum class Foo{BAR};
bool operator==(Foo a, Foo b)
{
return (int)a & (int)b;
}
int main(int argc, char *argv[])
{
Foo::BAR==Foo::BAR;
return 0;
}
MSVC fails with the following error:
>main.cpp(10): error C2593: 'operator ==' is ambiguous
>main.cpp(3): note: could be 'bool operator ==(Foo,Foo)'
>main.cpp(10): note: while trying to match the argument list '(Foo, Foo)'
Any insight would be great, I've been scratching my head about this all day.
My version of MSVC is 14.0 however I've tested it online with version Version 19.00.23506 and the same error appears.
The error does not apear with version 19.11.25331.0 however.
Compiler bug then?
For enumerations, there's a built-in comparison operator. When you define yours, the built-in is supposed to be hidden automatically.
[over.built/1]
The candidate operator functions that represent the built-in operators
defined in Clause [expr] are specified in this subclause. These
candidate functions participate in the operator overload resolution
process as described in [over.match.oper] and are used for no other
purpose. [ Note: Because built-in operators take only operands with
non-class type, and operator overload resolution occurs only when an
operand expression originally has class or enumeration type, operator
overload resolution can resolve to a built-in operator only when an
operand has a class type that has a user-defined conversion to a
non-class type appropriate for the operator, or when an operand has an
enumeration type that can be converted to a type appropriate for the
operator. Also note that some of the candidate operator functions
given in this subclause are more permissive than the built-in
operators themselves. As described in [over.match.oper], after a
built-in operator is selected by overload resolution the expression is
subject to the requirements for the built-in operator given in Clause
[expr], and therefore to any additional semantic constraints given
there. If there is a user-written candidate with the same name and
parameter types as a built-in candidate operator function, the
built-in operator function is hidden and is not included in the set of
candidate functions. — end note ]
To answer your question, yes, it seems like a compiler bug.
Why with libstdc++ this works but with libc++ it fails? On gcc it also works:
bool b = std::cin;
You should add the language standard and compiler you compile with.
Until C++11, std::basic_ios had operator void*, since C++11 it has explicit operator bool instead.
The second one is explicit, meaning an implicit conversion like in your example cannot use it.
libstdc++ from the GNU project still unconditionally contains the pre-C++ conversion (Version 4.9.1):
operator void*() const
{ return this->fail() ? 0 : const_cast<basic_ios*>(this); }
The bug https://gcc.gnu.org/bugzilla/show_bug.cgi?id=56193 is RESOLVED-FIXED since 2014-09-24, so the next release should be corrected.
According to the C++ Standard (13.3.1.5 Initialization by conversion function, p.#1)
The conversion functions of S and its base classes are considered.
Those non-explicit conversion functions that are not hidden within S
and yield type T or a type that can be converted to type T via a
standard conversion sequence (13.3.3.1.1) are candidate functions. For
direct-initialization, those explicit conversion functions that are
not hidden within S and yield type T or a type that can be converted
to type T with a qualification conversion (4.4) are also candidate
functions.
Class std::basic_ios has explicit conversion function operator bool. As
this declaration
bool b = std::cin;
does not use the direct initialization (there is the copy initialization) then it seems it is a bug of the compiler, that is the declaration shall not be compiled.
This example seems to compile with VC10 and gcc (though my version of gcc is very old).
EDIT: R. Martinho Fernandez tried this on gcc 4.7 and the behaviour is still the same.
struct Base
{
operator double() const { return 0.0; }
};
struct foo
{
foo(const char* c) {}
};
struct Something : public Base
{
void operator[](const foo& f) {}
};
int main()
{
Something d;
d["32"];
return 0;
}
But clang complains:
test4.cpp:19:6: error: use of overloaded operator '[]' is ambiguous (with operand types 'Something' and 'const char [3]')
d["32"]
~^~~~~
test4.cpp:13:10: note: candidate function
void operator[](const foo& f) {}
^
test4.cpp:19:6: note: built-in candidate operator[](long, const char *)
d["32"]
^
test4.cpp:19:6: note: built-in candidate operator[](long, const restrict char *)
test4.cpp:19:6: note: built-in candidate operator[](long, const volatile char *)
test4.cpp:19:6: note: built-in candidate operator[](long, const volatile restrict char *)
The overload resolution is considering two possible functions from looking at this expression:
calling Something::operator[] (after a user defined conversion)
calling built in operator for const char* (think "32"[d]) (after a user defined conversion and standard conversion double to long).
If I had written d["32"] as d.operator[]("32"), then overload resolution won't even look at option 2, and clang will also compile fine.
EDIT: (clarification of questions)
This seems to be a complicated area in overload resolution, and because of that I'd appreciate very much answers that explain in detail the overload resolution in this case, and cite the standard (if there's some obscure/advanced likely to be unknown rule).
If clang is correct, I'm also interested in knowing why the two are ambiguous / one is not preferred over another. The answer likely would have to explain how overload resolution considers implicit conversions (both user defined and standard conversions) involved on the two candidates and why one is not better than the other.
Note: if operator double() is changed to operator bool(), all three (clang, vc, gcc) will refuse to compile with similar ambiguous error.
It should be easier to picture why the overload resolution is ambiguous by going through it step-by-step.
§13.5.5 [over.sub]
Thus, a subscripting expression x[y] is interpreted as x.operator[](y) for a class object x of type T if T::operator[](T1) exists and if the operator is selected as the best match function by the overload resolution mechanism (13.3.3).
Now, we first need an overload set. That's constructed according to §13.3.1 and contains member aswell as non-member functions. See this answer of mine for a more detailed explanation.
§13.3.1 [over.match.funcs]
p2 The set of candidate functions can contain both member and non-member functions to be resolved against the same argument list. So that argument and parameter lists are comparable within this heterogeneous set, a member function is considered to have an extra parameter, called the implicit object parameter, which represents the object for which the member function has been called. [...]
p3 Similarly, when appropriate, the context can construct an argument list that contains an implied object argument to denote the object to be operated on.
// abstract overload set (return types omitted since irrelevant)
f1(Something&, foo const&); // linked to Something::operator[](foo const&)
f2(std::ptrdiff_t, char const*); // linked to operator[](std::ptrdiff_t, char const*)
f3(char const*, std::ptrdiff_t); // linked to operator[](char const*, std::ptrdiff_t)
Then, an argument list is constructed:
// abstract argument list
(Something&, char const[3]) // 'Something&' is the implied object argument
And then the argument list is tested against every member of the overload set:
f1 -> identity match on argument 1, conversion required for argument 2
f2 -> conversion required for argument 1, conversion required for argument 2 (decay)
f3 -> argument 1 incompatible, argument 2 incompatible, discarded
Then, since we found out that there are implicit conversions required, we take a look at §13.3.3 [over.match.best] p1:
Define ICSi(F) as follows:
if F is a static member function, [...]; otherwise,
let ICSi(F) denote the implicit conversion sequence that converts the i-th argument in the list to the type of the i-th parameter of viable function F. 13.3.3.1 defines the implicit conversion sequences and 13.3.3.2 defines what it means for one implicit conversion sequence to be a better conversion sequence or worse conversion sequence than another.
Now let's construct those implicit conversion sequences for f1 and f2 in the overload set (§13.3.3.1):
ICS1(f1): 'Something&' -> 'Someting&', standard conversion sequence
ICS2(f1): 'char const[3]' -> 'foo const&', user-defined conversion sequence
ICS1(f2): 'Something&' -> 'std::ptrdiff_t', user-defined conversion sequence
ICS2(f2): 'char const[3]' -> 'char const*', standard conversion sequence
§13.3.3.2 [over.ics.rank] p2
a standard conversion sequence (13.3.3.1.1) is a better conversion sequence than a user-defined conversion sequence or an ellipsis conversion sequence.
So ICS1(f1) is better than ICS1(f2) and ICS2(f1) is worse than ICS2(f2).
Conversely, ICS1(f2) is worse than ICS1(f1) and ICS2(f2) is better than ICS2(f1).
§13.3.3 [over.match.best]
p1 (cont.) Given these definitions, a viable function F1 is defined to be a better function than another viable function F2 if for all arguments i, ICSi(F1) is not a worse conversion sequence than ICSi(F2), and then [...]
p2 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.
Well, f*ck. :) As such, Clang is correct in rejecting that code.
It seems there is no question that both Something::operator[](const foo& f) and the built-in operator[](long, const char *) are viable candidate functions (13.3.2) for overload resolution. The types of real arguments are Something and const char*, and implicit conversion sequences (ICFs) I think are:
for Something::operator[](const foo& f): (1-1) identity conversion, and (1-2) foo("32") through foo::foo(const char*);
for operator[](long, const char *): (2-1) long(double(d)) through Something::operator double() const (inherited from Base), and (2-2) identity conversion.
Now if we rank these ICFs according to (13.3.3.2), we can see that (1-1) is a better conversion than (2-1), and (1-2) is a worse conversion than (2-2). According to the definition in (13.3.3),
a viable function F1 is defined to be a better function than another viable function
F2 if for all arguments i, ICSi(F1) is not a worse conversion sequence than ICSi(F2), ...
Therefore, neither of the considered two candidate functions is better than the other one, and thus the call is ill-formed. I.e. Clang appears to be correct, and the code should not compile.
It would seem from 13.6 in the C++11 spec that clang is correct here:
13.6
Built-in operators
[over.built]
The candidate operator functions that represent the built-in operators defined in Clause 5 are specified in
this subclause. These candidate functions participate in the operator overload resolution process as described
in 13.3.1.2 and are used for no other purpose. [ Note: Because built-in operators take only operands with
non-class type, and operator overload resolution occurs only when an operand expression originally has class
or enumeration type, operator overload resolution can resolve to a built-in operator only when an operand
has a class type that has a user-defined conversion to a non-class type appropriate for the operator, or when
an operand has an enumeration type that can be converted to a type appropriate for the operator. Also note
that some of the candidate operator functions given in this subclause are more permissive than the built-in
operators themselves. As described in 13.3.1.2, after a built-in operator is selected by overload resolution
the expression is subject to the requirements for the built-in operator given in Clause 5, and therefore to
any additional semantic constraints given there. If there is a user-written candidate with the same name
and parameter types as a built-in candidate operator function, the built-in operator function is hidden and
is not included in the set of candidate functions. — end note ]
:
For every cv-qualified or cv-unqualified object type T there exist candidate operator functions of the form
T& operator[](T *, std::ptrdiff_t);
T& operator[](std::ptrdiff_t, T *);
edit
Once you get past which operator functions exist, this just becomes standard overload resolution as described by section 13.3 of the standard -- about 10 pages of details, but the gist of it is that for a function call to not be ambiguous, there needs to be a single function that is at least as good a match as all the possible, viable functions on every argument, and a better match than the others on at least one argument. There's a lot of spec detail on exactly what 'better' means, but it boils down to (in this case) a match not requiring any user-defined conversion operator or object constructor is better than one which does.
So in this case, there are two viable matches:
void Something::operator[](const foo& f)
operator[](long, const char *)
The first is a better match for the first argument, while the second is a better match for the second. So unless there's some other function that is better than both of these, its ambiguous.
That latter point is a possble workaround -- add:
void operator[](const char *a) { return (*this)[foo(a)]; }
to class Something