Even though I do not declare the variables unix and linux and even not include anything in my teste.c compiling with GCC on Linux I have:
Compilation
gcc teste.c -o teste
Execution
./teste
Output
Unix and Linux are worth 1
Code
main()
{
if(linux * unix)
puts("Unix e Linux valem 1");
}
Are they reserved words of the language? Why are the two worth 1?
These are not reserved words but predefined macros when the compiler is running in a Linux / Unix environment.
At the command line, try the following command to list all of 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
The ANSI C standard of 1989 introduced rules that require that the name of a macro be started by two underscores or by an underscore followed by a small letter.
You can say that gcc, by default, "does not respect" these same rules, meaning that you can not 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 your reply, you can change this behavior using the options:
gcc -std=c90 -pedantic
gcc -std=c99 -pedantic
gcc -std=c11 -pedantic
gcc is a multi-language compiler.
link
By default, GCC provides some extensions to the C language that on rare occasions conflicts with the C standard. See Extensions to the C Language Family . ...
The way you called it is a compiler for "GNUC89".
In "GNUC89" the unix and linux symbols are set to 1 in your implementation.
To use gcc as a C11 compiler (the latest version of Standard), invoke it as well
gcc -std=c11 -pedantic ...