c++ - 对内联函数有不同的定义是未定义的行为吗？

Question

最少的代码:

// --------inline.h--------
struct X { 
  static inline void foo ();
};   
#ifdef YES
inline void X::foo () { cout << "YES\n"; }
#else
inline void X::foo () { cout << "NO\n"; }
#endif

// --------file1.cpp--------
#define YES    // <---- 
#include"inline.h"
void fun1 ()
{
  X::foo();
}

// --------file2.cpp--------
#include"inline.h"
void fun2 ()
{
  X::foo();
}

如果我们调用fun1()和fun2()，那么它们将分别打印YES和NO ，这意味着它们引用相同 X::foo() 的不同函数体。

不管这是否应该编码，我的问题是:
这是明确定义还是未定义的行为？

Answer 1

是的，这是未定义的行为。

引用:

C++03 标准:

7.1.2 函数说明符 [dcl.fct.spec]
第 4 段:

An inline function shall be defined in every translation unit in which it is used and shall have exactly the same definition in every case (3.2). [Note: a call to the inline function may be encountered before its definition appears in the translation unit. ] If a function with external linkage is declared inline in one translation unit, it shall be declared inline in all translation units in which it appears; no diagnostic is required. An inline function with external linkage shall have the same address in all translation units. A static local variable in an extern inline function always refers to the same object. A string literal in an extern inline function is the same object in different translation units.

注意:3.2指的是一个定义规则，其中规定:

3.2 一条定义规则[basic.def.odr]
第 1 段:

No translation unit shall contain more than one definition of any variable, function, class type, enumeration type or template.