std::max与max宏 混淆解决办法（转）

Today I typed the following:

int t = (std::max)(timeout, lagtime);

Why did I put parentheses around std::max? Because windows.h defines (among other things) a max and a min macro. If you include windows.h the above code will not compile. For example the following:

#include "windows.h"
#include <algorithm>

void foo() {
    int i = 5;
    int j = 7;
    int x = std::max(i,j);
}

Will produce the following error with Visual Studio C++ 2005:

1>test.cpp(7) : error C2589: ‘(‘ : illegal token on right side of ‘::‘
1>test.cpp(7) : error C2143: syntax error : missing ‘;‘ before ‘::‘

There are a number of ways to work around windows.h defining these two macros.

• Use alternative names defined in windows.h.

  int x = _cpp_max(i,j);
  int y = _cpp_min(i,j);

  This is not portable; only works on Windows.

• Define NOMINMAX before including windows.h. This might break existing code that assumes NOMINMAX is not defined.
• Don‘t use std::min and std::max. Instead use the tertiary operator like so:

  int x = i > j ? i : j; // max(i,j)
  int y = i < j ? i : j; // min(i,j)

  This is portable but not as readable and more error prone.

• Use using statements to make the code portable:

  using std::min;
  using std::max;
  int x = max(i,j);
  int y = min(i,j);

  This works but requires two more lines of code. You could also just use ‘using namespace std;‘ but that might pull in more than you want.

• Use std::min<int> and std::max<int>

  int x = std::max<int>(i,j);
  int y = std::min<int>(i,j);

  This requires you to specify the type. However in some cases this actually helps. For example:

  int i = 5;
  unsigned int j = 7;
  int x = (std::max)(i,j);
  int y = (std::min)(i,j);

  Note the ‘unsigned‘. Generates the following errors:

  1>test.cpp(7) : error C2780: ‘const _Ty &std::max(const _Ty &,const _Ty &,_Pr)‘ :
  expects 3 arguments - 2 provided
  1>        c:\program files\microsoft visual studio 8\vc\include\xutility(3190) :
  see declaration of ‘std::max‘
  1>test.cpp(7) : error C2782: ‘const _Ty &std::max(const _Ty &,const _Ty &)‘ :
  template parameter ‘_Ty‘ is ambiguous
  1>        c:\program files\microsoft visual studio 8\vc\include\xutility(3182) :
  see declaration of ‘std::max‘
  1>        could be ‘unsigned int‘
  1>        or ‘int‘

  By explicitly specifying type via <int> you remove the ambiguity.

• Use (std::min) and (std::max)

  int i = 5;
  int j = 7;
  int x = (std::max)(i,j);
  int y = (std::min)(i,j);

  This works (as does the std::max<int>) because the C++ preprocessor requires ‘(‘ as the next preprocessing token following the macro name to preform the macro expansion.