Unix and linux are reserved words?

Question

Even if the variables are not declared unix and linux and even without including anything in my teste.c compiling with the GCC on Linux have:

Compilation

gcc teste.c -o teste

Execution

./teste

Exit

Unix and Linux are worth 1

Code

main()
{
        if(linux * unix)
                puts("Unix e Linux valem 1");
}

Are words reserved from language? Because both are worth 1?

Answer 1

Not reserved words but predefined macros when the compiler is running on a Linux/Unix environment.

Try, on the command line, to run the following command to list all these macros:

gcc -dM < /dev/null

The two words linux and Unix have the following definition:

gcc -dM < /dev/null | grep linux

#define __linux 1
#define __linux__ 1
#define __gnu_linux__ 1
#define linux 1

gcc -dM < /dev/null | grep unix 
#define __unix__ 1
#define __unix 1
#define unix 1

1989 ANSI C standard introduced rules that require a macro name to be started by two underscores or an underscore followed by a letter maisculas.

You can say that gcc by default "doesn’t respect" those same rules, meaning you can’t do something like

int main() {

   int unix = 10;

}

This will generate the following message:

error: expected identifier or '(' before numeric constant
  int unix = 10;

As @pmg indicated in his reply, you can change this behavior using the options:

gcc -std=c90 -pedantic
gcc -std=c99 -pedantic
gcc -std=c11 -pedantic

Answer 2

The gcc is a compiler for several languages.
https://gcc.gnu.org/onlinedocs/gcc-5.2.0/gcc/Standards.html#Standards

By default, GCC provides some Extensions to the C language that on Rare occasions Conflict with the C standard. See Extensions to the C Language Family. ...

The way you called it he’s a compiler for "GNUC89".

In "GNUC89" the symbols unix and linux are defined as 1 in your implementation.

To use gcc as the C11 compiler (the latest version of the Standard), invoke it like this

gcc -std=c11 -pedantic ...