#include <string>
void foo(int x, short y, int z) { std::cout << "normal int" << std::endl; } //F1
void foo(double x, int y, double z) { std::cout << "normal double" << std::endl; } //F2
int main()
{
short x = 2;
foo(5.0, x, 8.0);
}
Based on function overloading rules,
F1( Exact = 1, Promote = 0, Convert = 2 ) and
F2( Exact = 2, Promote = 1, Convert = 0 ). Shouldn't F2 be called? Why is the call ambiguous?
Overload resolution can get complicated. But here it's fairly straightforward. I'll rewrite the function prototypes to remove the third argument, since it doesn't affect the analysis.
void foo(int, short);
void foo(double, int);
double d = 1.0;
short s = 2;
f(d, s); // ambiguous
The rule is that you look at each argument, one at a time, and determine which function has the "best" conversion for that argument. If one function has the best conversion for every argument, that's the one that's called. If there's no function that's best for every argument, the call is ambiguous.
For the first argument, the type of the argument is double. Calling the first version of foo requires a conversion from double to int; calling the second version of foo requires no conversion; it's an exact match. So for the first argument, the second version of foo has the best conversion.
For the second argument, the type of the argument is short. Calling the first version of foo requires no conversion; it's an exact match. Calling the second version of foo requires a promotion from short to int. So for the second argument, the first version of foo has the best conversion.
Since the first version of foo has the best match for the second argument and the second version of foo has the best match for the first argument, there is no function with the best match on all arguments, and the call is ambiguous.
Overload resolution ranks the viable candidate functions by comparing how each argument of the call matches the corresponding parameter of the candidates. Moreover, for one candidate to be considered better than another, the better candidate cannot have any of its parameters be a worse match than the
corresponding parameter in the other candidate.
Now, let's apply this to your given example snippet.
void foo(int x, short y, int z);
void foo(double x, int y, double z) ;
int main()
{
short x = 2;
foo(5.0, x, 8.0);
}
In the above snippet, For the case of first overloaded foo, the second argument matches exactly with the type of the second parameter. But the first and third argument have type double while the corresponding parameters have type int and so the first and the third argument requires a conversion to int in this case.
On the other hand, for the case of second overloaded foo, the first and third argument matches exactly with the first and third parameter respectively. But in this case, the second argument is of type short while the second parameter is of type int. So a promotion to int is needed for the second argument.
Result 1
Now, according to the highlighted part at the beginning of my answer, for the first candidate to be considered a better match than the second candidate, none of the first candidate's parameter can be a worse match than the corresponding parameter in the second candidate. But we already discussed that the first and third parameter of first overloaded candidate are worse match than the corresponding parameters in the second overloaded candidate since they require conversion to int. Thus, the first overloaded candidate is not a better match than the second overloaded candidate.
Result 2
Similarly, for the second candidate to be considered a better match than the first candidate, none of the second candidate's parameter can be a worse match than the corresponding parameter in the first candidate. But here also we already discussed that the second parameter in the second candidate is worse match than the corresponding parameter in the first candidate since it requires a promotion. Thus, the second candidate is not a better match than the first candidate.
Conclusion
Combining both of the results above, we get to the conclusion that none of the candidates is better than the other. Hence the call is ambiguous.
The ranking of conversion sequences matters only when comparing ranking of the conversion sequences applied to the same argument.
If one argument's conversion rank is better in the first overload than in the second one and the other way around for another argument's conversion rank, then neither overload is considered better than the other and the overload resolution is ambiguous.
This applies here. The first and third argument have a better conversion rank (exact rather than conversion) in the second overload, while the second argument has a better conversion rank in the first overload (exact rather than promotion).
Related
#include <iostream>
using namespace std;
void x(int a,int b){
cout<<"int int"<<endl;
}
void x(char a,char b){
cout<<"char char"<<endl;
}
int main() {
int a =2;char c ='a';
x(a,c);
return 0;
}
call to 'x' is ambiguous in apple clang compiler, why?
for x(int,int), first argument is a direct match and second is a promotion
for x(char, char) first argument is a standard conversion as I know and also according to this answer-> https://stackoverflow.com/a/28184631/13023201
And promotion should be preferred over std conversion, then x(int,int) should be called. Then why is this ambiguous??
Looking at how cppreference describes the overload resolution process:
Best viable function
For each pair of viable function F1 and F2, the implicit conversion sequences from the i-th argument to i-th parameter are ranked to determine which one is better (except the first argument, the implicit object argument for static member functions has no effect on the ranking)
F1 is determined to be a better function than F2 if implicit conversions for all arguments of F1 are not worse than the implicit conversions for all arguments of F2, and
...
x(int, int) is a better match than x(char, char) for the first argument, since int to int is an exact match and int to char is a conversion. But x(int, int) is a worse match for the second argument, because it's a promotion instead of an exact match. So neither function is a better match than the other and the overload cannot be resolved.
I'm trying to understand the overloading resolution method.
Why is this ambiguous:
void func(double, int, int, double) {}
void func(int, double, double, double) {}
void main()
{
func(1, 2, 3, 4);
}
but this isn't?
void func(int, int, int, double) {}
void func(int, double, double, double) {}
void main()
{
func(1, 2, 3, 4);
}
In the first case there are 2 exact parameters matches and 2 conversions against 1 exact match and 3 conversions, and in the second case there are 3 exact matches and 1 conversion against 1 exact matches and 3 conversions.
So why is one ambiguous and one is not? What is the logic here?
The overload resolution rules only define a partial order on the set of all matches - if an overload F1 is not a better match than F2, it does not imply that F2 is a better match than F1. The exact partial order can be thought of as comparing two points in k dimensions, where the number of arguments is k. Lets define this partial order on points in k-dim space - (x_1, x_2,..., x_k) < (y_1, y_2,..., y_k) if x_i <= y_i for all i and x_j < y_j for at least one j. This is exactly the partial order on candidate non-template functions defined by the standard.
Lets look at your examples :
void func(double, int, int, double) {}
vvv vvv vvv
better better equal
void func(int, double, double, double) {}
vvv vvv
better equal
So neither overload is strictly better than the other.
In your second example:
void func(int, int, int, double) {}
vvv vvv vvv vvv
equal better better equal
void func(int, double, double, double) {}
vvv
equal
Now, the first overload is better than the second in all but one argument AND is never worse than the second. Thus, there is no ambiguity - the partial order does indeed declare the first one better.
(The above description does not consider function templates. You can find more details at cppreference.)
The wording from the standard (§[over.match.best]/1) is:
[...] 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.
[...] 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
— for some argument j, ICSj(F1) is a better conversion sequence than ICSj(F2)
In your first case, the two functions fail the first test. For the first argument, the first function (taking double) has a worse conversion sequence than the second. For the second argument, the second function has a worse conversion sequence than the first (again, the int has to be promoted to double in one case, but not the other).
Therefore, neither function passes the first rule, and the call is ambiguous.
Between the second pair of functions, every argument to the the first function has at least as good of a conversion as the matching argument to the second function. We then go on to the second rule, and find that there is at least one argument (two, as a matter of fact) for which the first function has a better conversion (identity instead of promotion) than the second.
Therefore, the first function is a better match, and will be selected.
Ambiguity is determined by the ranking:
Exact match: no conversion required, lvalue-to-rvalue conversion, qualification conversion, user-defined conversion of class type to the same class
Promotion: integral promotion, floating-point promotion
Conversion: integral conversion, floating-point conversion, floating-integral conversion, pointer conversion, pointer-to-member conversion, boolean conversion, user-defined conversion of a derived class to its base
Exact match wins vs Promotion which wins vs Conversion.
In the example:
void func(int, bool, float, int){cout << "int,bool,float,int" << endl;}
void func(int, bool, int, int){cout << "int,int,int,int" << endl;}
int main()
{
func(1,1,3.4,4);
}
Argument 1(1) is an exact match on both
Argument 2(1) is an exact match on both
Argument 3(3.4) can be converted into float and int - Ambiguity Neither is better.
Argument 4(4) is an exact match on both
But if we did this: func(1,1,3.4f,4); (3.4f) is now an exact match!
void func(int, bool, float, int) then wins the battle.
I don't understand what happens here
class A{};
class B : A {};
void func(A&, bool){}
void func(B&, double){}
int main(void)
{
B b;
A a;
bool bo;
double d;
func(b, bo);
}
When compiling, Visual 2010 gives me this error on line func(b, bo);
2 overloads have similar conversions
could be 'void func(B &,double)'
or 'void func(A &,bool)'
while trying to match the argument list '(B, bool)'
I don't understand why the bool parameter isn't enough to resolve the overload.
I've seen this question, and as pointed in the accepted answer, bool should prefer the bool overload. In my case, I see that first parameter isn't enough to choose the good function, but why the second parameter doesn't solve the ambiguity?
The overloading rules are a bit more complicated than you might guess. You look at each argument separately and pick the best match for that argument. Then if there is exactly one overload that provides the best match for every argument, that's the one that's called. In the example, the best match for the first argument is the second version of func, because it requires only a conversion of B to B&; the other version of func requires converting B to B& and then B& to A&. For the second argument, the first version of func is the best match, because it requires no conversions. The first version has the best match for the second argument, but it does not have the best match for the first argument, so it is not considered. Similarly, the second version has the best match for the first argument, but it does not have the best match for the second argument, so it is not considered. Now there are no versions of func left, and the call fails.
Overload resolution rules are even more complicated than Pete Becker wrote.
For each overload of f, the compiler counts not only the number of parameters for which conversion is required, but the rank of conversion.
Rank 1
No conversions required
Lvalue-to-rvalue conversion
Array-to-pointer conversion
Function-to-pointer conversion
Qualification conversion
Rank 2
Integral promotions
Floating point promotions
Rank 3
Integral conversions
Floating point conversions
Floating-integral conversions
Pointer conversions
Pointer to member conversions
Boolean conversions
Assuming that all candidates are non-template functions, a function wins if and only if it has a parameter which rank is better than the rank of the same parameter in other candidates and the ranks for the other parameters are not worse.
Now let's have a look at the OP case.
func(A&, bool): conversion B&->A& (rank 3) for the 1st parameter,
exact match (rank 1) for the 2nd parameter.
func(B&, double): exact match (rank 1) for the 1st parameter, conversion bool->double (rank 3) for the 2nd parameter.
Conclusion: noone wins.
Consider I have the following minimal code:
#include <boost/type_traits.hpp>
template<typename ptr_t>
struct TData
{
typedef typename boost::remove_extent<ptr_t>::type value_type;
ptr_t data;
value_type & operator [] ( size_t id ) { return data[id]; }
operator ptr_t & () { return data; }
};
int main( int argc, char ** argv )
{
TData<float[100][100]> t;
t[1][1] = 5;
return 0;
}
GNU C++ gives me the error:
test.cpp: In function 'int main(int, char**)':
test.cpp:16: error: ISO C++ says that these are ambiguous, even though the worst conversion for the first is better than the worst conversion for second:
test.cpp:9: note: candidate 1: typename boost::remove_extent<ptr_t>::type& TData<ptr_t>::operator[](size_t) [with ptr_t = float [100][100]]
test.cpp:16: note: candidate 2: operator[](float (*)[100], int) <built-in>
My questions are:
Why GNU C++ gives the error, but Intel C++ compiler is not?
Why changing operator[] to the following leads to compiling without errors?
value_type & operator [] ( int id ) { return data[id]; }
Links to the C++ Standard are appreciated.
As I can see here are two conversion paths:
(1)int to size_t and (2)operator[](size_t).
(1)operator ptr_t&(), (2)int to size_t and (3)build-in operator[](size_t).
It's actually quite straight forward. For t[1], overload resolution has these candidates:
Candidate 1 (builtin: 13.6/13) (T being some arbitrary object type):
Parameter list: (T*, ptrdiff_t)
Candidate 2 (your operator)
Parameter list: (TData<float[100][100]>&, something unsigned)
The argument list is given by 13.3.1.2/6:
The set of candidate functions for overload resolution is the union of the member candidates, the non-member candidates, and the built-in candidates. The argument list contains all of the operands of the operator.
Argument list: (TData<float[100][100]>, int)
You see that the first argument matches the first parameter of Candidate 2 exactly. But it needs a user defined conversion for the first parameter of Candidate 1. So for the first parameter, the second candidate wins.
You also see that the outcome of the second position depends. Let's make some assumptions and see what we get:
ptrdiff_t is int: The first candidate wins, because it has an exact match, while the second candidate requires an integral conversion.
ptrdiff_t is long: Neither candidate wins, because both require an integral conversion.
Now, 13.3.3/1 says
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.
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 ... for some argument j, ICSj(F1) is a better conversion sequence than ICSj(F2), or, if not that ...
For our first assumption, we don't get an overall winner, because Candidate 2 wins for the first parameter, and Candidate 1 wins for the second parameter. I call it the criss-cross. For our second assumption, the Candidate 2 wins overall, because neither parameter had a worse conversion, but the first parameter had a better conversion.
For the first assumption, it does not matter that the integral conversion (int to unsigned) in the second parameter is less of an evil than the user defined conversion of the other candidate in the first parameter. In the criss-cross, rules are crude.
That last point might still confuse you, because of all the fuss around, so let's make an example
void f(int, int) { }
void f(long, char) { }
int main() { f(0, 'a'); }
This gives you the same confusing GCC warning (which, I remember, was actually confusing the hell out of me when I first received it some years ago), because 0 converts to long worse than 'a' to int - yet you get an ambiguity, because you are in a criss-cross situation.
With the expression:
t[1][1] = 5;
The compiler must focus on the left hand side to determine what goes there, so the = 5; is ignored until the lhs is resolved. Leaving us with the expression: t[1][1], which represents two operations, with the second one operating on the result from the first one, so the compiler must only take into account the first part of the expression: t[1].The actual type is (TData&)[(int)]
The call does not match exactly any functions, as operator[] for TData is defined as taking a size_t argument, so to be able to use it the compiler would have to convert 1 from int to size_t with an implicit conversion. That is the first choice. Now, another possible path is applying user defined conversion to convert TData<float[100][100]> into float[100][100].
The int to size_t conversion is an integral conversion and is ranked as Conversion in Table 9 of the standard, as is the user defined conversion from TData<float[100][100]> to float[100][100] conversion according to §13.3.3.1.2/4. The conversion from float [100][100]& to float (*)[100] is ranked as Exact Match in Table 9. The compiler is not allowed to choose from those two conversion sequences.
Q1: Not all compilers adhere to the standard in the same way. It is quite common to find out that in some specific cases a compiler will perform differently than the others. In this case, the g++ implementors decided to whine about the standard not allowing the compiler to choose, while the Intel implementors probably just silently applied their preferred conversion.
Q2: When you change the signature of the user defined operator[], the argument matches exactly the passed in type. t[1] is a perfect match for t.operator[](1) with no conversions whatsoever, so the compiler must follow that path.
I don't know what's the exact answer, but...
Because of this operator:
operator ptr_t & () { return data; }
there exist already built-in [] operator (array subscription) which accepts size_t as index. So we have two [] operators, the built-in and defined by you. Booth accepts size_t so this is considered as illegal overload probably.
//EDIT
this should work as you intended
template<typename ptr_t>
struct TData
{
ptr_t data;
operator ptr_t & () { return data; }
};
It seems to me that with
t[1][1] = 5;
the compiler has to choose between.
value_type & operator [] ( size_t id ) { return data[id]; }
which would match if the int literal were to be converted to size_t, or
operator ptr_t & () { return data; }
followed by normal array indexing, in which case the type of the index matches exactly.
As to the error, it seems GCC as a compiler extension would like to choose the first overload for you, and you are compiling with the -pedantic and/or -Werror flag which forces it to stick to the word of the standard.
(I'm not in a -pedantic mood, so no quotes from the standard, especially on this topic.)
I have tried to show the two candidates for the expression t[1][1]. These are both of equal RANK (CONVERSION). Hence ambiguity
I think the catch here is that the built-in [] operator as per 13.6/13 is defined as
T& operator[](T*, ptrdiff_t);
On my system ptrdiff_t is defined as 'int' (does that explain x64 behavior?)
template<typename ptr_t>
struct TData
{
typedef typename boost::remove_extent<ptr_t>::type value_type;
ptr_t data;
value_type & operator [] ( size_t id ) { return data[id]; }
operator ptr_t & () { return data; }
};
typedef float (&ATYPE) [100][100];
int main( int argc, char ** argv )
{
TData<float[100][100]> t;
t[size_t(1)][size_t(1)] = 5; // note the cast. This works now. No ambiguity as operator[] is preferred over built-in operator
t[1][1] = 5; // error, as per the logic given below for Candidate 1 and Candidate 2
// Candidate 1 (CONVERSION rank)
// User defined conversion from 'TData' to float array
(t.operator[](1))[1] = 5;
// Candidate 2 (CONVERSION rank)
// User defined conversion from 'TData' to ATYPE
(t.operator ATYPE())[1][1] = 6;
return 0;
}
EDIT:
Here is what I think:
For candidate 1 (operator []) the conversion sequence S1 is
User defined conversion - Standard Conversion (int to size_t)
For candidate 2, the conversion sequence S2 is
User defined conversion -> int to ptrdiff_t (for first argument) -> int to ptrdiff_t (for second argument)
The conversion sequence S1 is a subset of S2 and is supposed to be better. But here is the catch...
Here the below quote from Standard should help.
$13.3.3.2/3 states - Standard
conversion sequence S1 is a better
conversion sequence than standard
conversion sequence S2 if — S1 is a
proper subsequence of S2 (comparing
the conversion sequences in the
canonical form defined by 13.3.3.1.1,
excluding any Lvalue Transformation;
the identity conversion sequence is
considered to be a subsequence of any
non-identity conversion sequence) or,
if not that...
$13.3.3.2 states- " 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 and
if the second standard conversion
sequence of U1 is better than the
second standard conversion sequence of
U2."
Here the first part of the and condition "if they contain the same user-defined conversion function or constructor" does not hold good. So, even if the second part of the and condition "if the second standard conversion sequence of U1 is better than the second standard conversion sequence of U2." holds good, neither S1 nor S2 is preferred over the other.
That's why gcc's phantom error message "ISO C++ says that these are ambiguous, even though the worst conversion for the first is better than the worst conversion for the second"
This explains the ambiguity quiet well IMHO
Overload resolution is a headache. But since you stumbled on a fix (eliminate conversion of the index operand to operator[]) which is too specific to the example (literals are type int but most variables you'll be using aren't), maybe you can generalize it:
template< typename IT>
typename boost::enable_if< typename boost::is_integral< IT >::type, value_type & >::type
operator [] ( IT id ) { return data[id]; }
Unfortunately I can't test this because GCC 4.2.1 and 4.5 accept your example without complaint under --pedantic. Which really raises the question whether it's a compiler bug or not.
Also, once I eliminated the Boost dependency, it passed Comeau.
Given the code below, why is the foo(T*) function selected ?
If I remove it (the foo(T*)) the code still compiles and works correctly, but G++ v4.4.0 (and probably other compilers as well) will generate two foo() functions: one for char[4] and one for char[7].
#include <iostream>
using namespace std;
template< typename T >
void foo( const T& )
{
cout << "foo(const T&)" << endl;
}
template< typename T >
void foo( T* )
{
cout << "foo(T*)" << endl;
}
int main()
{
foo( "bar" );
foo( "foobar" );
return 0;
}
Formally, when comparing conversion sequences, lvalue transformations are ignored. Conversions are grouped into several categories, like qualification adjustment (T* -> T const*), lvalue transformation (int[N] -> int*, void() -> void(*)()), and others.
The only difference between your two candidates is an lvalue transformation. String literals are arrays that convert to pointers. The first candidate accepts the array by reference, and thus won't need an lvalue transformation. The second candidate requires an lvalue transformation.
So, if there are two candidates that both function template specializations are equally viable by looking only at the conversions, then the rule is that the more specialized one is chosen by doing partial ordering of the two.
Let's compare the two by looking at their signature of their function parameter list
void(T const&);
void(T*);
If we choose some unique type Q for the first parameter list and try to match against the second parameter list, we are matching Q against T*. This will fail, since Q is not a pointer. Thus, the second is at least as specialized as the first.
If we do the other way around, we match Q* against T const&. The reference is dropped and toplevel qualifiers are ignored, and the remaining T becomes Q*. This is an exact match for the purpose of partial ordering, and thus deduction of the transformed parameter list of the second against the first candidate succeeds. Since the other direction (against the second) didn't succeed, the second candidate is more specialized than the first - and in consequence, overload resolution will prefer the second, if there would otherwise be an ambiguity.
At 13.3.3.2/3:
Standard conversion sequence S1 is a better conversion sequence than standard conversion sequence S2 if [...]
S1 is a proper subsequence of S2 (comparing the conversion sequences in the canonical form
defined by 13.3.3.1.1, excluding any Lvalue Transformation; the identity conversion sequence is considered to be a subsequence of any non-identity conversion sequence) or, if not that [...]
Then 13.3.3/1
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.
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 [...]
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 14.5.5.2, or, if not that, [...]
Finally, here is the table of implicit conversions that may participate in an standard conversion sequence at 13.3.3.1.1/3.
Conversion sequences http://img259.imageshack.us/img259/851/convs.png
The full answer is quite technical.
First, string literals have char const[N] type.
Then there is an implicit conversion from char const[N] to char const*.
So both your template function match, one using reference binding, one using the implicit conversion. When they are alone, both your template functions are able to handle the calls, but when they are both present, we have to explain why the second foo (instantiated with T=char const[N]) is a better match than the first (instantiated with T=char). If you look at the overloading rules (as given by litb), the choice between
void foo(char const (&x)[4));
and
void foo(char const* x);
is ambigous (the rules are quite complicated but you can check by writing non template functions with such signatures and see that the compiler complains). In that case, the choice is made to the second one because that one is more specialized (again the rules for this partial ordering are complicated, but in this case it is because you can pass a char const[N] to a char const* but not a char const* to a char const[N] in the same way as void bar(char const*) is more specialized than void bar(char*) because you can pass a char* to a char const* but not vise-versa).
Based on overload resolution rules (Appendix B of C++ Templates: The Complete Guide has a good overview), string literals (const char []) are closer to T* than T&, because the compiler makes no distinction between char[] and char*, so T* is the closest match (const T* would be an exact match).
In fact, if you could add:
template<typename T>
void foo(const T[] a)
(which you can't), your compiler would tell you that this function is a redefinition of:
template<typename T>
void foo(const T* a)
Cause " " is a char*, which fits perfectly to foo(T*) function. When you remove this, the compiler will try to make it work with foo(T&), which requires you to pass reference to char array that contains the string.
Compiler can't generate one function that would receive reference to char, as you are passing whole array, so it has to dereference it.