In Xcode6, what's the "compiler default" for C++ language dialect.
I am using a C++ new feature std:max(a,b,c)
if I use "Compiler Default", it failed to compile.
When I changed to "C++11 or GNUC++11", it compiles fine.
I am wondering if compiler default is C++98?
I ran below code to get - GNU C++ 98.
#include <iostream>
int main()
{
//gnu mode
#ifndef __STRICT_ANSI__
std::cout << "GNU - ";
#endif
// C++ iso standard
#if __cplusplus == 199711L
std::cout << "C++98" << std::endl;
#elif __cplusplus == 201103L
std::cout << "C++11" << std::endl;
#elif __cplusplus > 201103L
std::cout << "C++14" << std::endl;
#endif
}
Macros chosen
__cplusplus - From gcc online documentation
Depending on the language standard selected, the value of the macro is 199711L, as mandated by the 1998 C++ standard; 201103L, per the 2011 C++ standard; an unspecified value strictly larger than 201103L for the experimental languages enabled by -std=c++1y and -std=gnu++1y.
__STRICT_ANSI__ - From clang user manual
Differences between all c* and gnu* modes => c* modes define
__STRICT_ANSI__
As a side note, __STRICT_ANSI__ for GNU standard differentiation could also be found from this SO answer
$ g++ -E -dM -std=c++11 -x c++ /dev/null >b
$ g++ -E -dM -std=gnu++11 -x c++ /dev/null >a
$ diff -u a b
--- a 2014-12-19 12:27:11.000000000 +0530
+++ b 2014-12-19 12:27:05.000000000 +0530
@@ -144,6 +144,7 @@
#define __STDC_UTF_16__ 1
#define __STDC_UTF_32__ 1
#define __STDC__ 1
+#define __STRICT_ANSI__ 1
#define __UINTMAX_TYPE__ long unsigned int
#define __USER_LABEL_PREFIX__ _
#define __VERSION__ "4.2.1 Compatible Apple LLVM 6.0 (clang-600.0.54)"
