std::min and std::max definitions

I have the following statements in my stdafx.h file:

using std::min;
using std::max;

My compiler is telling me that min and max are not members of std.

What am I missing?

I am using VS 6.0 and have included the following:

Who is Participating?
itsmeandnobodyelseConnect With a Mentor Commented:
>>>> The rest of the code base uses the std::min/max functions
Unfortunately, VC6 was released prior to C++ Standard, so its STL implementation is not compliant in many points.  

In the MSDN comment they told that there is a conflict with with (macro) definitions in WINDEF.H. You can get rid of that macro by adding NOMINMAX to the preprocessor macros of your project settings. Then in your cpp files you could add at the top (below inlcude of stdafx.h if using PCH).

#ifndef min
#define min_defined
#define min std::_cpp_min

#ifndef max
#define max_defined
#define max std::_cpp_max

and et the bottom of the cpp:

#ifdef min_defined
#undef min
#ifdef max_defined
#undef max

Of course you could put both sequences into a header file which can be included then.

Isn't it ugly? Maybe a better way is to copy the template definition of _ccp_min and _cpp_max, put them to a new header file and rename the function names to min and max accordingly. Put both function to the namespace std. If you uccessfully suppressed any other occurrence fo min and max you can include that header what should make valid the above usage clauses.

Regards, Alex


#include <xutility>
using std::_cpp_min;
using std::_cpp_max;
MSDN that came with VC6 says:

template<class T>
    const T& min(const T& x, const T& y);
template<class T, class Pred>
    const T& min(const T& x, const T& y, Pred pr);
The first template function returns y if y < x. Otherwise it returns x. T need supply only a single-argument constructor and a destructor.

The second template function behaves the same, except that it replaces operator<(X, Y) with pr(X, Y).

To avoid conflicts with min and max in WINDEF.H, use _MIN and _MAX instead. These macros evaluate to _cpp_min and _cpp_max, respectively.
END Microsoft-Specific

So you either use jkr's solution or take _MIN and _MAX .

Regards, Alex
StanChartAuthor Commented:
I'm actually merging some code right now and what I previously had was:

using namespace std;

If I replace that with using std::_cpp_min/max, I get other errors like auto_ptr not defined, etc.
If I include both, I get "none of the 2 overloads have a best conversion".

The rest of the code base uses the std::min/max functions so I'd like to stick with those since it is not my code that i would be modifying.

And if you use

using namespace std;

it works?
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