Compiler Error C2666

'identifier' : number overloads have similar conversions

An overloaded function or operator is ambiguous. Formal parameter lists may be too similar for the compiler to resolve the ambiguity. To resolve this error, explicitly cast one or more of the actual parameters.

Examples

The following sample generates C2666:

// C2666.cpp
struct complex {
   complex(double);
};

void h(int,complex);
void h(double, double);

int main() {
   h(3,4);   // C2666
}

This error can be generated as a result of compiler conformance work that was done for Visual Studio 2019 version 16.1:

• A conversion that promotes an enumeration whose underlying type is fixed to its underlying type is better than one that promotes to the promoted underlying type, if the two are different.

The following example demonstrates how compiler behavior changes in Visual Studio 2019 version 16.1 and later versions:

#include <type_traits>

enum E : unsigned char { e };

int f(unsigned int)
{
    return 1;
}

int f(unsigned char)
{
    return 2;
}

struct A {};
struct B : public A {};

int f(unsigned int, const B&)
{
    return 3;
}

int f(unsigned char, const A&)
{
    return 4;
}

int main()
{
    // Calls f(unsigned char) in 16.1 and later. Called f(unsigned int) in earlier versions.
    // The conversion from 'E' to the fixed underlying type 'unsigned char' is better than the
    // conversion from 'E' to the promoted type 'unsigned int'.
    f(e);
  
    // Error C2666. This call is ambiguous, but previously called f(unsigned int, const B&). 
    f(e, B{});
}

This error can also be generated as a result of compiler conformance work that was done for Visual Studio .NET 2003:

• binary operators and user-defined conversions to pointer types

• qualification conversion is not the same as identity conversion

For the binary operators <, >, <=, and >=, a passed parameter is now implicitly converted to the type of the operand if the parameter's type defines a user-defined conversion operator to convert to the type of the operand. There is now potential for ambiguity.

For code that is valid in both the Visual Studio .NET 2003 and Visual Studio .NET versions of Visual C++, call the class operator explicitly using function syntax.

// C2666b.cpp
#include <string.h>
#include <stdio.h>

struct T
{
    T( const T& copy )
    {
        m_str = copy.m_str;
    }

    T( const char* str )
    {
        int iSize = (strlen( str )+ 1);
        m_str = new char[ iSize ];
        if (m_str)
            strcpy_s( m_str, iSize, str );
    }

    bool operator<( const T& RHS )
    {
        return m_str < RHS.m_str;
    }

    operator char*() const
    {
        return m_str;
    }

    char* m_str;
};

int main()
{
    T str1( "ABCD" );
    const char* str2 = "DEFG";

    // Error - Ambiguous call to operator<()
    // Trying to convert str1 to char* and then call
    // operator<( const char*, const char* )?
    //  OR
    // trying to convert str2 to T and then call
    // T::operator<( const T& )?

    if( str1 < str2 )   // C2666

    if ( str1.operator < ( str2 ) )   // Treat str2 as type T
        printf_s("str1.operator < ( str2 )\n");

    if ( str1.operator char*() < str2 )   // Treat str1 as type char*
        printf_s("str1.operator char*() < str2\n");
}

The following sample generates C2666

// C2666c.cpp
// compile with: /c

enum E
{
    E_A,   E_B
};

class A
{
    int h(const E e) const {return 0; }
    int h(const int i) { return 1; }
    // Uncomment the following line to resolve.
    // int h(const E e) { return 0; }

    void Test()
    {
        h(E_A);   // C2666
        h((const int) E_A);
        h((int) E_A);
    }
};