c – 对内联函数有不同的定义是否是一个未定义的行为？

最小代码：

// --------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,这意味着它们是引用相同X :: foo()的不同函数体.

无论是否应该编码,我的问题是：
这是一个明确定义还是未定义的行为？

解决方法

是的是未定义的行为.

参考：

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指一定义规则,其中规定：

3.2一个定义规则[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.