isnan函数 linux,为什么isnan含糊不清以及如何避免它？

既然isnan可以是宏(在C++ 98中)或在命名空间std中定义的函数(在C++ 11中),这个简单的例子说明了编写在两种情况下都有效的代码的明显(并且可能是天真的)方式.

#include

int main() {

double x = 0;

using namespace std;

isnan(x);

}

但是,编译它会在GCC(使用-std = c ++ 11)和Clang中产生错误:

test.cc: In function ‘int main()’:

test.cc:6:10: error: call of overloaded ‘isnan(double&)’ is ambiguous

isnan(x);

^

test.cc:6:10: note: candidates are:

In file included from /usr/include/features.h:374:0,

from /usr/include/x86_64-linux-gnu/c++/4.8/bits/os_defines.h:39,

from /usr/include/x86_64-linux-gnu/c++/4.8/bits/c++config.h:426,

from /usr/include/c++/4.8/cmath:41,

from test.cc:1:

/usr/include/x86_64-linux-gnu/bits/mathcalls.h:234:1: note: int isnan(double)

__MATHDECL_1 (int,isnan,, (_Mdouble_ __value)) __attribute__ ((__const__));

^

In file included from test.cc:1:0:

/usr/include/c++/4.8/cmath:626:3: note: constexpr bool std::isnan(long double)

isnan(long double __x)

^

/usr/include/c++/4.8/cmath:622:3: note: constexpr bool std::isnan(double)

isnan(double __x)

^

/usr/include/c++/4.8/cmath:618:3: note: constexpr bool std::isnan(float)

isnan(float __x)

^

为什么在C++ 11中这是不明确的,以及如何使它适用于C++ 98和C++ 11,最好不需要太多的条件编译？