Does extern C with C++ avoid undefined behavior that is legal in C but not C++?

Question

If you use extern C it with C++ files, does that allow defined C behavior that is undefined in C++?

blah.h

 extern "C"
 {
      struct x {
           int blah;
           char buf[];
      };

      char * get_buf(struct x * base);
      struct x * make_struct(int blah, int size);
 }

some_random.cpp

 #include "blah.h"

 ...

 x * data=make_struct(7, 12);
 std::strcpy(get_buf(data), "hello");

Is using the defined behavior in C's flexible array member, defined behavior when used this way?

Note: Since get_buf() make_struct() only use a pointer to struct x, why is struct x even defined in blah.h? — chux
– chux, Commented Aug 7, 2015 at 16:07
I'm rolling this question back to the prior version. If you have a new question, you are welcome to ask it, but your last edit appears to be an attempt to invalidate all the answers you received. — Drew Dormann
– Drew Dormann, Commented Aug 7, 2015 at 16:08
@DrewDormann I made this clearly extern "C" and asked the linkage question independantly — Glenn Teitelbaum
– Glenn Teitelbaum, Commented Aug 7, 2015 at 16:40
If this was not language-lawyer, I'd point out that #ifdef guarding char buf[]; out of existence makes the code work. — Yakk - Adam Nevraumont
– Yakk - Adam Nevraumont, Commented Aug 8, 2015 at 4:54

Keith Thompson · Accepted Answer · 2015-08-07 18:15:21Z

19

Flexible array members are a standard feature of C, starting with the 1999 standard. They do not exist in C++.

Your code is not valid C++. Wrapping it in extern "C" doesn't change that. A conforming C++ compiler must at least warn about it, and arguably should reject it.

It happens that g++ implements C-style flexible array members as an extension to C++. That's perfectly legitimate (compilers are allowed to implement extensions), but its use is not portable. Its behavior, like that of any language extension, is defined by the compiler, not by the language.

If you compile it with g++ -pedantic, you'll get a warning:

c.cpp:5:21: warning: ISO C++ forbids zero-size array ‘buf’ [-Wpedantic]
            char buf[];
                     ^

If you want to use C-style flexible array members in a C++ program without relying on a compiler-specific extension, you can compile your C code as C and link it into your C++ program. You can't make the type with the flexible array member visible to your C++ code, but you can use it internally in the C code, and perhaps provide access to it in your C++ code via an opaque pointer. See the C++ FAQ for information about mixing C and C++ within the same program. (Or you can just use the g++ extension, at the cost of not being able to compile your code with other compilers.)

(I'm assuming that you're using g++. Some other compilers probably implement similar extensions.)

edited Aug 7, 2015 at 18:15

answered Aug 7, 2015 at 15:53

Keith Thompson

266k46 gold badges447 silver badges668 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Kenny Ostrom Over a year ago

You may need to hide struct x's actual definition from the c++ header, and treat it as an opaque pointer except when you access it through the defined C functions.

Keith Thompson Over a year ago

@KennyOstrom: Yes, that's what I was suggesting (though not as clearly as you did). I'll update my answer.

Quentin · Accepted Answer · 2015-08-07 15:51:46Z

11

No.

extern "C" is only a linkage specification, so that enclosed symbols can be linked to from C. It does not switch your compiler to "C mode" for a section of code.

answered Aug 7, 2015 at 15:51

Quentin

63.5k7 gold badges139 silver badges199 bronze badges

2 Comments

Quentin Over a year ago

@GlennTeitelbaum It doesn't matter : the problem is that flexible array members don't exist in C++. Even if only a C compiler got to see it used (it's not the case, you try to std::copy to it), the C++ compiler would choke on its declaration alone.

Keith Thompson Over a year ago

True, flexible array members don't exist in standard C++, but they do exist in the g++ dialect of C++, as a compiler-specific extension.

user4815162342 · Accepted Answer · 2015-08-07 15:51:56Z

10

An extern "C" declaration only affects linkage of external functions, so that name mangling is not performed. It doesn't mean that the functions will be compiled using the rules of the C language instead of C++. In other words, including the code in some_random.cpp will not make its behavior defined.

answered Aug 7, 2015 at 15:51

user4815162342

159k22 gold badges350 silver badges418 bronze badges

Comments

Jon Chesterfield · Accepted Answer · 2015-08-07 15:52:08Z

-5

It's just ascii until such time as the compiler runs. So the preprocessor will paste the files together, then the compiler will treat the result as whatever language you specified.

It isn't a C flexible array member, it's a text file. Undefined behaviour remains.

answered Aug 7, 2015 at 15:52

Jon Chesterfield

2,3511 gold badge22 silver badges32 bronze badges

2 Comments

Lightness Races in Orbit Over a year ago

Missed the point; the question is whether extern "C" makes valid C / illegal C++ into a sort of pseudo legal C++.

Jon Chesterfield Over a year ago

It didn't occur to me that extern C could be expected to achieve so much, while determining what language a text file is processed as seems to throw people. Thank you for your comment.

Collectives™ on Stack Overflow

Does extern C with C++ avoid undefined behavior that is legal in C but not C++?

4 Answers 4

2 Comments

2 Comments

Comments

2 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

2 Comments

2 Comments

Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related