Re: Compilation failure involving template operator

=?ISO-8859-1?Q?Daniel_Kr=FCgler?= <>
Sun, 4 Jul 2010 16:03:03 CST
On 4 Jul., 10:49, Edward Diener <eldie...@tropicsoft.invalid> wrote:

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)

XTestChar xtest;
YTestTemplate<char> ytest;

void AFunction()
  ytest *= xtest;



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 ?

The allowed operator overloads behave like other function
templates (as explained in detail within [over.match.oper]).

You have decided for a non-member operator overload
of operator*=. The way you defined this template has
the effect that the compiler is supposed to deduce
the template parameter T for both it's T-dependent
arguments. Doing so, it finds T=char for the first and
T=XTestChar for the second. Since this is a function
template, no conversions are tried.

You have different ways to solve this problem, e.g.:

1) Use a member function overload like this:

template <class T> struct YTestTemplate {
 YTestTemplate &
 operator *=(T second) {
  return *this;

This is a non-template member-function of
YTestTemplate, therefore the compiler looks
only at the member candidates of YTestTemplate
and finds only one. Note that in this case
the T of the member *= is fixed and no
T deduction of this argument is done. Now
the compiler looks for conversions to this T
and successfully finds the conversion function,
leading to success.

2) Use a non-member overload like as you started
with, but this time ensure that the second
argument is in a non-deduced context, e.g. like

template<class T> struct identity { typedef T type; };

template <class T>
YTestTemplate<T> &
operator *= (YTestTemplate<T> & first, typename identity<T>::type
  return first;

This time your original program should be well-formed,
because the compiler does no longer attempt to deduce
two (potentially different) parameters types with T.

HTH & Greetings from Bremen,

Daniel Kr?gler

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