What does the g++ -D flag do?
compilation, flags, g++, linux
Solution
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.
Problem
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