Write a program that will print "C" if compiled as an (ANSI) C program, and "C++" if compiled as a C++ program

1. Abuse C++ automatic typedefs

(Note that the struct needs to be declared in an inner scope so that it takes precedence over the outer name in C++.)

#include <stdio.h>

int main(void)
{
    char x;

    {
        struct x { char dummy[2]; };
        printf("%s\n", sizeof (x) == 1 ? "C" : "C++");
    }
}

A similar version that doesn't rely on the ambiguity between sizeof (type) and sizeof (variable), using only types:

#include <stdio.h>

int main(void)
{
    typedef char t;

    {
        struct t { char dummy[2]; };
        printf("%s\n", sizeof (t) == 1 ? "C" : "C++");
    }
}

2. Abuse C++ struct/class equivalence, automatic typedefs, and automatically-generated default constructors

#include <stdio.h>

int isC = 0;
void Foo() { isC = 1; }

int main(void)
{
    struct Foo { int dummy; };
    Foo();
    printf("%s\n", isC ? "C" : "C++");
}

3. Abuse nested struct declarations in C

Also see Symbol clashing of inner and outer structs, C++ vs C

#include <stdio.h>

int main(void)
{
    typedef struct inner { int dummy; } t;

    {
        struct outer { struct inner { t dummy[2]; } dummy; };
        printf("%s\n",
               sizeof (struct inner) == sizeof (t)
               ? "C++"
               : "C");
    }
}

4. Abuse // comments

This won't work with C99 or with C89 compilers that support // as an extension.

#include <stdio.h>

int main(void)
{
    printf("%s\n",
           0 //* */
           +1
           ? "C++"
           : "C");
}

or alternatively:

    printf("%s\n",
           1 //* */ 2
           ? "C++"
           : "C");

5. sizeof differences with char literals

Note that this isn't guaranteed to be portable since it's possible that some hypothetical platform could use bytes with more than 8 bits, in which case sizeof(char) could be the same as sizeof(int). (Also see Can sizeof(int) ever be 1 on a hosted implementation?)

#include <stdio.h>

int main(void)
{
    printf("%s\n", sizeof 'a' == 1 ? "C++" : "C");
}

6. Abuse differences in when lvalue⇒rvalue conversions are performed

This is based off of the 5.16, 5.17, 5.18 example in the ISO C++03 standard, and it works in gcc but not in MSVC (possibly due to a compiler bug?).

#include <stdio.h>

int main(void)
{
    void* array[2];
    printf("%s\n",
           (sizeof (((void) 0), array) / sizeof (void*) == 1)
           ? "C"
           : "C++");
}

7. Abuse differences in the way C and C++'s grammars parse the ternary operator

This one isn't strictly legal, but some compilers are lax.

#include <stdio.h>

int main(void)
{
    int isCPP = 1;
    printf("%s\n", (1 ? isCPP : isCPP = 0) ? "C++" : "C");
}

(You also could check for the __cplusplus preprocessor macro (or various other macros), but I think that doesn't follow the spirit of the question.)

I have implementations for all of these at: http://www.taenarum.com/csua/fun-with-c/c-or-cpp.c


We had to do a similar assignment at school. We were not allowed to use preprocessor (except for #include of course). The following code uses the fact that in C, type names and structure names form separate namespaces whereas in C++ they don't.

#include <stdio.h>
typedef int X;
int main()
{
    struct X { int ch[2]; };
    if (sizeof(X) != sizeof(struct X))
        printf("C\n");
    else
        printf("C++\n");
}

Simple enough.

#include <stdio.h>
int main(int argc, char ** argv) {
#ifdef __cplusplus
printf("C++\n");
#else
printf("C\n");
#endif
return 0;
}

Or is there a requirement to do this without the official standard?

Tags:

C++

C

Compiler Specific

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