Identificateur de configuration win64 dans Qmake

Question

Y at-il un identifiant « win64 » dans les fichiers de projet QMake? Qt Qmake avancé la documentation ne mentionne pas autre qu'Unix / macx / win32.

Jusqu'à présent, je l'ai essayé d'utiliser:

win32:message("using win32")
win64:message("using win64")
amd64:message("using amd64")

Le résultat est toujours "à l'aide win32".

Dois-je utiliser un fichier de projet distinct pour les projets x32 et x64, donc ils compilerions contre les bibliothèques correctes? Est-il un autre moyen d'identifier entre 32 bits et les environnements 64 bits?

Answer 1

Je fais comme ça

win32 {

    ## Windows common build here

    !contains(QMAKE_TARGET.arch, x86_64) {
        message("x86 build")

        ## Windows x86 (32bit) specific build here

    } else {
        message("x86_64 build")

        ## Windows x64 (64bit) specific build here

    }
}

Answer 2

Since Qt5 you can use QT_ARCH to detect whether your configuration is 32 or 64. When the target is 32-bit, that returns i386 and in case of a 64-bit target it has the value of x86_64. So it can be used like:

contains(QT_ARCH, i386) {
    message("32-bit")
} else {
    message("64-bit")
}

Answer 3

UPDATE: since very recently, Qt has a way of doing this transparently and easily, without manual hassle:

win32-g++:contains(QMAKE_HOST.arch, x86_64):{
    do something
}

Source: the brand new Qt Dev FAQ

Answer 4

I've figured out one way to do it.

Qt allows you to pass arbitrary config parameters which you can use to separate the targets.

By having a conditional config in your project file:

CONFIG(myX64, myX64|myX32) {
    LIBPATH += C:\Coding\MSSDK60A\Lib\x64
} else {
    LIBPATH += C:\Coding\MSSDK60A\Lib
}

and passing that custom config to qmake with

qmake CONFIG+=myX64

you get the wanted result.

Answer 5

No, but you can create and use a new mkspec, I think qmake also defines a platform identifier named after the current mkspec. Why do you need to test for 64 bit?

Reed