Why does the C preprocessor interpret the word "linux" as the constant "1"?

Question

Why does the C preprocessor in GCC interpret the word `linux` (small letters) as the constant `1`? test.c: #include <stdio.h> int main(void) { int linux = 5; return 0; } Result of `$ gcc -E test.c` (stop after the preprocessing stage): .... int main(void) { int 1 = 5; return 0; } Which -of course- yields an error. (BTW: There is no `#define linux` in the stdio.h file.)

Answer 1

The short answer is that you need to use a different variable name.

Longer answer: This is a hold-over from the ancient days of C, pre-ANSI. It allowed the preprocessor to determine at compile time which system is being used to compile the program and on which system the final program will be run.

unix, vax, linux, and a few others will be reserved words until the next standard of C, but they should be avoided.

The new standards state that reserved words must begin with two underscores "_", or a single underscore followed by an uppercase letter. This is intended to free up all valid names for programmers to use as variables.

Now another part of this that you need to understand is that gcc doesn't conform to standards unless you tell it to, and even then it's still a bit off.

Here is a link to the gcc standards that explains this point in further detail: https://gcc.gnu.org/onlinedocs/gcc-4.8.1/gcc/Standards.html

Answer 2

gcc defines a collection of predefined macros that are system specific. While compilers are moving away frm them, there are a number of macros -- like linux -- that are still around for historical reasons.

If you compile with -ansi or with -std=c99 that will suppress the old fashioned macros.

See https://gcc.gnu.org/onlinedocs/cpp/System-specific-Predefined-Macros.html#System-specific-Predefined-Macros for more on this

Answer 3

Works fine in Bloodshed Dev C/C++

returns the zero to the DOS shell in 1/10 of a second...

I must say I have never seen void being passed to main...

make sure your syntax is correct...

#include <stdio.h>

int main( void)

{

int linux=1;

return 0;

}

Answer 4

You're right: There is no #define linux in the header file. There are certain system-specific macros that are injected by the preprocessor. From the C preprocessor documentation:

The C standard requires that all system-specific macros be part of the reserved namespace. All names which begin with two underscores, or an underscore and a capital letter, are reserved for the compiler and library to use as they wish. 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.

When the -ansi option, or any -std option that requests strict conformance, is given to the compiler, all the system-specific predefined macros outside the reserved namespace are suppressed. The parallel macros, inside the reserved namespace, remain defined.

We are slowly phasing out all predefined macros which are outside the reserved namespace. You should never use them in new programs, and we encourage you to correct older code to use the parallel macros whenever you find it.

Out of curiosity, I ran `cpp -dM foo.c` on the Windows subsystem for Linux, and I saw both `linux` and `unix` defined. I was able to compile your code using `gcc -ansi foo.c`.

Here is the page of the C preprocessor documentation: https://gcc.gnu.org/onlinedocs/cpp/System-specific-Predefined-Macros.html#System-specific-Predefined-Macros