Is "unix" restricted keyword in C?

unix is not a identifier reserved by the Standard.

If you compile with -std=c89 or -std=c99 the gcc compiler will accept the program as you expected.

From gcc manual ( https://gcc.gnu.org/onlinedocs/cpp/System-specific-Predefined-Macros.html ), the emphasis is mine.

... However, historically system-specific macros have had names with no special prefix; for instance, it is common to find unix defined on Unix systems. For all such macros, GCC provides a parallel macro with two underscores added at the beginning and the end. If unix is defined, __unix__ will be defined too. There will never be more than two underscores; the parallel of _mips is __mips__.

It is not a keyword.

It is a predefined macro to identify the type of system. On Unix and Unix like systems it is defined to be 1.

To disable this use the -ansi option:

In C mode, this is equivalent to -std=c89. In C++ mode, it is equivalent to -std=c++98. This turns off certain features of GCC that are incompatible with ISO C90 (when compiling C code), or of standard C++ (when compiling C++ code), such as the "asm" and "typeof" keywords, and predefined macros such as "unix" and "vax" that identify the type of system you are using. It also enables the undesirable and rarely used ISO trigraph feature. For the C compiler, it disables recognition of C++ style // comments as well as the "inline" keyword.

unix is one of the defines the preprocessor uses in gcc to get a list of defs use

gcc -dM -E  -x c /dev/null

(-dM tells gcc to debugdump the defs -E tells it to stop after prepreocessing and -x c /dev/null tells him to pretend /dev/null is a c file)

Run your code through the preprocessor to find out what the compiler is actually seeing:

gcc -E unix.c

Then see if your variable unix is preserved or converted by the preprocessor.