What does the g++ -D flag do?

Question

I am looking at a CFLAGS of -

CFLAGS=-g -w -D LINUX -O3 -fpermissive

in a Makefile. What does the -D flag do? I see on the man page that

-D name
    Predefine name as a macro, with definition 1. 

but I don't know how to interpret that. My interpretation is...its making LINUX a macro and only doing -03 and -fpermissive when in a linux environment. Is that right? If not, then what? Thanks for any help

Solution 1

It is equivalent to adding the statement #define LINUX 1 in the source code of the file that is being compiled. It does not have any effect on other compilation flags. The reason for this is it's an easy way to enable #ifdef statements in the code. So you can have code that says:

#ifdef LINUX
   foo;
#endif

It will only be enabled if that macro is enabled which you can control with the -D flag. So it is an easy way to enable/disable conditional compilation statements at compile time without editing the source file.

Solution 2

It doesn't have anything to do with -O3. Basically, it means the same as

#define LINUX 1

at the beginning of the compiled file.

Comments

[Basile Starynkevitch]

Often, the -D is glued to the following (defined) name, e.g. -DLINUX or -DFOO=BAR

[jpm]

I would +1, but the interpretation is only correct up to a point. The -D flag doesn't affect the following flags (e.g. making them conditional or some such crazy talk), and that's a fairly important distinction.

[Gabriel Southern]

@jpm that's a good point I had not interpreted the question that way when I first read it, but now I see you are correct. I edited my answer to clarify that point.

[jpm]

-O3 sets the compiler's optimization level to 3.