tell if make is running on windows or linux

Question

Is there a way to know in makefiles if GNU make is running on a linux OS or a windows OS?

I've built a bash script that generates a makefile for building my app and it works fine on my Debian machine. I want to try to build it on MinGW/MSYS, but the problem is that I have to build and run some test programs that check errors in source code, and to run it on Windows, I must add the .exe suffix.

Answer 1

UPDATE
Please read this similar but better answer:
https://stackoverflow.com/a/14777895/938111

---

make (and gcc) can be easily installed on MS-Windows using Cygwin or MinGW.

As @ldigas says, make can detect the platform using UNAME:=$(shell uname) (the command uname is also installed by Cygwin or MinGW installer).

Below, I provide a complete example based on make (and gcc) to explain how to build a shared library: *.so or *.dll depending on the platform.

The example is basic/simple to be easily understandable :-)

Let's see the five files:

 ├── app
 │   └── Makefile
 │   └── main.c
 └── lib
     └── Makefile
     └── hello.h
     └── hello.c

The Makefiles

app/Makefile

app.exe: main.o
        gcc -o $@ $^ -L../lib -lhello
        # '-o $@'    => output file => $@ = the target file (app.exe)
        # '   $^'    => no options => Link all depended files 
        #            => $^ = main.o and other if any
        # '-L../lib' => look for libraries in directory ../lib
        # '-lhello   => use shared library hello (libhello.so or hello.dll)

%.o: %.c
        gcc -o $@ -c $< -I ../lib
        # '-o $@'     => output file => $@ = the target file (main.o)
        # '-c $<'     => COMPILE the first depended file (main.c)
        # '-I ../lib' => look for headers (*.h) in directory ../lib

clean:
        rm -f *.o *.so *.dll *.exe

lib/Makefile

UNAME := $(shell uname)

ifeq ($(UNAME), Linux)
TARGET = libhello.so
else
TARGET = hello.dll
endif

$(TARGET): hello.o
        gcc  -o $@  $^  -shared
        # '-o $@'    => output file => $@ = libhello.so or hello.dll
        # '   $^'    => no options => Link all depended files => $^ = hello.o
        # '-shared'  => generate shared library

%.o: %.c
        gcc  -o $@  -c $<  -fPIC
        # '-o $@' => output file => $@ = the target file (hello.o)
        # '-c $<' => compile the first depended file (hello.c)
        # '-fPIC' => Position-Independent Code (required for shared lib)

clean:
        rm -f *.o *.so *.dll *.exe

The source code

app/main.c

#include "hello.h" //hello()
#include <stdio.h> //puts()

int main()
{
    const char* str = hello();
    puts(str);
}

lib/hello.h

#ifndef __HELLO_H__
#define __HELLO_H__

const char* hello();

#endif

lib/hello.c

#include "hello.h"

const char* hello()
{
    return "hello";
}

The build

Fix Makefiles copy (replace leading spaces by tabulation).

> sed  -i  's/^  */\t/'  */Makefile

The make command is the same on both platforms. This is the output on MS-Windows (removed unnecessary lines).

> cd lib
> make clean
> make
gcc  -o hello.o  -c hello.c  -fPIC
gcc  -o hello.dll  hello.o  -shared
> cd ../app
> make clean
> make
gcc -o main.o -c main.c -I ../lib
gcc -o app.exe main.o -L../lib -lhello

The run

The application requires to know where is the shared library.

On MS-Windows, the simple/basic/stupid way is to copy the library where the application is:

> cp -v lib/hello.dll app
`lib/hello.dll' -> `app/hello.dll'

On Linux, use the LD_LIBRARY_PATH environment variable:

> export LD_LIBRARY_PATH=lib

The run command line and output are the same on both platforms:

> app/app.exe
hello