How to tell gcc to compile for Windows ?

I wrote a C program in which valid instructions are used depending on the platform on which the program is to be started: Windows or Linux. In the C code I use #ifdef USE_WIN to say that that piece of code goes for the Windows platform, unlike #else ....
How do I go about passing USE_WIN inside the Makefile via the command line with make ?

I forgot, I use gcc as a compiler, both in Windows and Linux.

There are some predefined symbols for Windows: _WIN32 (this exists on mingw and VC++), __WINNT__

Pre-defined Compiler Macros / Wiki / Home

Tim S

As a compiler I use gcc of MinGW in Windows 10 and gcc in Linux.
So far, I know too, but I would like to go command line with make, telling which platform to use in the build. Example: make WIN_USE and use WIN_USE in gcc which it defines in source code (using #ifdef WIN_USE do this #else do that).

I'll explain. I have the following source code main.c:

Code:

---

#include <stdio.h>

int main(int argc, char *argv[]) {
#ifdef LINUX
    printf("Ciao Linux\n");
#else
    printf("Ciao Windows\n");
#endif
   
    return 0;
}

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The Makefile is:

Code:

---

all: uno

uno:
    gcc -D $(CPLAT) -c main.c
    gcc -o test main.o

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When I start make I have to pass the CPLAT variable (which is either LINUX or WIN) to the Makefile, thus compiling the source code main.c obtaining a different result depending on the variable passed CPLAT (which will affect the #ifdef LINUX ... #else in main .c).
How should I do ?

I got there now. The solution is very simple:

Code:

---

CPLAT = LINUX

all: uno

uno:
    gcc -D $(CPLAT) -c main.c
    gcc -o test main.o

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then from the command line I type:

Quote:

---

make CPLAT=LINUX

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or

Quote:

---

make CPLAT=WIN

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so I get respectively with ./test:

Quote:

---

Ciao Linux

---

,

Quote:

---

Ciao Windows

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