内联函数如何在C ++中工作？

Question

now I have 2 c++ source files: test9.cpp test10.cpp, they both have a inline function with the same name.

test9.cpp:

1 #include <iostream>
2 using namespace std;
3 void test();
4 inline void f1()
5 {
6     cout << "inline f1 in test9.cpp" << endl;
7 }   
8 int main()
9 {
10     f1();
11     test();
12     return 0;
13 } 

test10.cpp:

1 #include <iostream>
2 using namespace std;
3 inline void f1()
4 {
5     cout << "inline f1 in test10.cpp" << endl;
6 }   
7 void test()
8 {
9     f1();
10 } 

now compile them with g++: g++ test9.cpp test10.cpp ./a.out I get the following result:

inline f1 in test9.cpp
inline f1 in test9.cpp

what's wrong? I thought it would be: "inline f1 in test9.cpp inline f1 in test10.cpp" who can tell me why? how does the g++ compiler treats the inline function?

Answer 1

While the compiler will allow you (nay, require you!) to redefine functions marked inline , the default of external linkage still applies, so you are breaking the One Definition Rule. This results in undefined behaviour and the outcome that you are seeing.

[C++11: 7.1.2/4]: An inline function shall be defined in every translation unit in which it is odr-used and shall have exactly the same definition in every case (3.2). [..]