在 C、C++ 中检测 Windows 或 Linux

我正在写一个跨平台的程序.我希望这一个程序同时在Windows和Linux下运行，所以我有两个不同的代码段用于这两个平台.如果操作系统是Windows，我希望运行第一个代码段；如果是Linux，那么我希望运行第二个代码段.

因此，我编写了以下代码，但在Windows和Linux上构建时出现错误.我应该怎么做才能解决这个问题呢？

#ifdef __unix__                    /* __unix__ is usually defined by compilers targeting Unix systems */

    #define OS_Windows 0
    #include <unistd.h>
    #include <stdlib.h>
    #include <stdio.h>
    #include <string.h>

#elif defined(_WIN32) || defined(WIN32)     /* _Win32 is usually defined by compilers targeting 32 or   64 bit Windows systems */

    #define OS_Windows 1
    #include <windows.h>
    #include <stdio.h>
    #include <tchar.h>
    #define DIV 1048576
    #define WIDTH 7

#endif

int main(int argc, char *argv[])
{
    if(OS_Windows)
    {
        MEMORYSTATUSEX statex;
        statex.dwLength = sizeof (statex);
        GlobalMemoryStatusEx (&statex);

        _tprintf (TEXT("There is  %*ld %% of memory in use.\n"),
                    WIDTH, statex.dwMemoryLoad);

    }

    else if(!OS_Windows) // if OS is unix

    {
        char cmd[30];
        int flag = 0;
        FILE *fp;
        char line[130];
        int memTotal, memFree, memUsed;

        flag=0;
        memcpy (cmd,"\0",30);
        sprintf(cmd,"free -t -m|grep Total");
        fp = popen(cmd, "r");
        while ( fgets( line, sizeof line, fp))
        {
            flag++;
            sscanf(line,"%*s %d %d %d",&TotalMem, &TotalUsed, &TotalFree);
        }
        pclose(fp);

        if(flag)
            printf("TotalMem:%d -- TotalUsed:%d -- TotalFree:%d\n",TotalMem,TotalUsed,TotalFree);
        else
            printf("not found\n");

    }

    return 0;
}