unknown type name C vs C++

Question

I tried to create a struct that holds a pointer to it's type:

#include <stdio.h>
struct test{
    test* t;
};
 int main(){
    return 0;
 }

while compiled with gcc, this code produced an error:

:5:2: error: unknown type name 'test'

but while compiled on g++ it went just fine.

So what I want to understand is:

1) What causes this difference? I thought that if gcc uses one-pass compilation and g++ uses multipass that could explain it, but as I understood, this is not true.

2) How can I avoid this error and define a struct with a pointer to it's own type? (I don't want to use a void* or use casting unless there is no other option)

@Viatorus tried that, but when dereferencing that afterwards I got "incomplete type" error — CIsForCookies, Jul 18 '16 at 06:21
See the answer here: http://stackoverflow.com/questions/612328/difference-between-struct-and-typedef-struct-in-c — Dúthomhas, Jul 18 '16 at 06:22
C and C++ are different but related languages, and you should have no expectation that a C++ program is also a valid C program, although the opposite is usually true. — Paul Hankin, Jul 18 '16 at 06:26
C and C++ are different languages. Please choose one tag, which is appropriate for your question. — CiaPan, Jul 18 '16 at 06:28
@CiaPan OP actually ask for the difference between these languages. — πάντα ῥεῖ, Jul 18 '16 at 06:31

score 6 · Accepted Answer · answered Jul 18 '16 at 06:22

6

You have to write struct before test* t to define a pointer to our struct:

struct test* t;

That is one different between C and C++.

answered Jul 18 '16 at 06:22

Viatorus

1,804
1
18
41

Magisch · Answer 2 · 2016-07-18T06:26:56.007

2

Your syntax is incorrect in "C". You have to use struct <structname>* <varname>; in format.

In c++, you can omit the "struct" in front of it. In C however, you have to write it to define a pointer to your struct.

Fixed code:

#include <stdio.h>
typedef struct test{
   struct test* t;
} test;
 int main(){
    return 0;
 }

Live demo

edited Jul 18 '16 at 06:26

answered Jul 18 '16 at 06:24

Magisch

7,312
9
36
52

score 0 · Answer 3 · answered Jul 18 '16 at 06:30

0

You need to typedef struct test test ; before you can use it in that way.

This should work just fine:

#include <stdio.h>
struct test{
   struct test* t;
};
int main(){
    return 0;
}

Or you can try:

#include <stdio.h>
typedef struct test test;
struct test{
   test* t;
};
int main(){
    return 0;
}

Your problem is gcc is a C compiler and takes types literally. If you use a c++ compiler, it may not care. c++ simplifies the types for you C does not. Be careful of this fact. C code will compile in a c++ compiler but c++ will not compile in a c compiler.

answered Jul 18 '16 at 06:30

haelmic

541
4
18

1

"*C code will compile in a c++ compiler*" - most C code isn't valid C++. – melpomene Jul 18 '16 at 06:31
It depends if you are using strict mode or not. Depending on the c++ compiler to ensure backwards compatibility the compiler will just do it. – haelmic Jul 18 '16 at 06:35

unknown type name C vs C++

3 Answers3