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Subject: Re: Why does overload resolution fail in this simple case?
From: Hani Deek (hani_deek_at_[hidden])
Date: 2020-12-30 13:36:07


Sorry, my last message was a mistake. The code is accepted by all compilers. The issue was only in the enum declaration.

________________________________
 The following example produces quite conflicting results https://godbolt.org/z/PExfM1.
It is accepted by msvc and icc, but rejected by gcc and clang. Because of these different results, in addition to the point made by Lénárd Szolnoki in his last message, I still do not know the correct answer to this whole issue.

enum A;
enum B;

struct S
{
    operator A() &{ return A{}; }
    operator B() &&{ return B{}; }
};

void foo(A) {}
void foo(B) {}

int main()
{
    foo(S{});
}

________________________________
Hani Deek via Std-Discussion <std-discussion_at_[hidden]<mailto:std-discussion_at_[hidden]>>

Hello,

With MSVC, the following code compiles. The char overload is selected by overload resolution.

struct S
{
        operator int() const & { return 0; }
        operator char() && { return 0; }
};

void foo(int) {}
void foo(char) {}

int main()
{
        foo(S{}); //OK, calls 'void foo(char)'.
}

However, the following code won't compile.

struct S
{
        int i = 0;
        operator const int &() const & { return i; }
        operator int &&() && { return (int &&)(i); }
};

void foo(const int &) {}
void foo(int &&) {}

int main()
{
        foo(S{}); //error C2668: 'foo': ambiguous call to overloaded function
                  //message : could be 'void foo(int &&)'
                  //message : or 'void foo(const int &)'
}

What is exactly the difference that makes the second sample fail to compile?

Given that 'operator int &&() &&' is an exact match to the conversion required to call 'void foo(int &&)', I expected the compiler to select it. It is strange if the C++ rules will not allow the compiler to select a conversion function that exactly matches the required conversion, both in terms of the provided argument 'S &&' and the result of conversion 'int &&'.

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