Re: Compilation failure involving template operator

The code below surprisingly for me fails compilation on Comeau C++ and
VC++ 9:

----------------------------------------------------------

struct XTestChar
{
operator char() const { return c; }
char c;
};

template <class T> struct YTestTemplate { };

template <class T>
YTestTemplate<T> &
operator *= (YTestTemplate<T> & first,T second)
{
return(first);
}

XTestChar xtest;
YTestTemplate<char> ytest;

void AFunction()
{
ytest *= xtest;
}

------------------------------------------------------------

Comeau C++ says:

"ComeauTest.c", line 21: error: no operator "*=" matches these operands
operand types are: YTestTemplate<char> *= XTestChar
ytest *= xtest;
^
1 error detected in the compilation of "ComeauTest.c".

Visual C++ says:

error C2782: 'YTestTemplate<T> &operator *=(YTestTemplate<T> &,T)' :
template parameter 'T' is ambiguous
see declaration of 'operator *='
could be 'XTestChar'
or 'char'
error C2676: binary '*=' : 'YTestTemplate<T>' does not define this
operator or a conversion to a type acceptable to the predefined operator
with
[
T=char
]

I would have thought that the C++ compiler would always deduce T in the
operator*= to be 'char', from the fact that 'ytest' is clearly
'YTestTemplate<char>' and then use XTestChar's char conversion operator
to pass a 'char' to the operator *= function second parameter. Why is
this not allowed ? Are templated operators not allowed to do conversions
when a type is deduced ?

If I have, in place of my template a normal class and a non-template
operator using that class:

struct YTestClass { };

YTestClass &
operator *= (YTestClass & first,char second)
{
return(first);
}

then

XTestChar xtest;
YTestClass ytestc;

void AFunction()
{
ytestc *= xtest;
}

compiles with no error.