Re: Stack overflow using boost::operators

From:
Yechezkel Mett <ymett.on.usenet@gmail.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Sun, 20 Dec 2009 08:48:34 CST
Message-ID:
<1b60d20d-2543-495e-ac66-b542278e65fd@j14g2000yqm.googlegroups.com>
On Dec 17, 3:32 pm, ManicQin <manic...@gmail.com> wrote:

Hello all,

while using boost::operators I bump into a stack overflow.

....

Using my debugger (visual studio 9 express) I jumped into
the operator.hpp and followed the sequence of the program,
for a reason I cannot understand in the first error line
which examples a T <= T operator I move from
less_than_comparable1::operator<=(const T& x, const T& y)
to
less_than_comparable2::operator<(const U& x, const T& y)

....

#include "boost/lexical_cast.hpp"
#include "boost/operators.hpp"

#include <iostream>

using namespace boost;
using namespace std;

....

template <class T>
class FieldOp : public T ,
  boost::ordered_field_operators<FieldOp<T>,
     boost::ordered_field_operators<FieldOp<T>,T,
                boost::ordered_field_operators<FieldOp<T>,typename T::FieldType
        > > >
{
public:

        FieldOp(typename T::FieldType const& val) : T(val) {}
        FieldOp(T const& val) : T(val.Data()) {}
        FieldOp(FieldOp<T> const& val) : T(val.Data()) {}

        bool operator > (typename T::FieldType const& val)
        {
                return (Data() > val);
        }
        bool operator > (typename T const& val)
        {
                return (Data() > val.Data());
        }
        bool operator > (typename FieldOp<T> const& val)
        {
                return (Data() > val.Data());
        }

        bool operator < (typename T::FieldType const& val)
        {
                return Data() < val;
        }
        bool operator < (typename T const& val)
        {
                return Data() < val.Data();
        }
        bool operator < (typename FieldOp<T> const& val)
        {
                return Data() < val.Data();
        }


These functions should be const.

What's happening is that the boost defined
operator>(const FieldOp<IField<int>>&,const FieldOp<IField<int>>&)
takes your object by const reference, and then tries to call operator<
with the arguments swapped. Since your operator< isn't const it can't
be called on a const reference. Since const FieldOp<IField<int>>&
converts automatically to const IField<int>& due to a derived-to-base
conversion, you instead end up in the boost defined
operator<(const IField<int>&,const FieldOp<IField<int>>&)
which tries to call operator> with the arguments swapped. Since your
operator> isn't const it can't be called on a const reference. Since
IField<int> converts automatically to const FieldOp<IField<int>>& via
a converting constructor, you instead end up in the boost defined
operator>(const FieldOp<IField<int>>&,const FieldOp<IField<int>>&)
which is where we were before. And so on and so forth, until the stack
overflows.

Incidentally it's generally recommended to make your operators free
functions.

Yechezkel Mett

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