当对象类型是模板参数时，有没有办法将模板参数传递给对象上的函数?

举例说明:

struct MyFunc {

    template <size_t N>
    void doIt() {
        cout << N << endl;
    }

};

template <typename Func>
struct Pass123ToTemplateFunc {

    static void pass(Func f) {
        f.doIt<123>(); // <-- Error on compile; is there a way to express this?
    }

};

int main() {

    Pass123ToTemplateFunc<MyFunc>::pass(MyFunc());

    return 0;

}

这几乎纯粹是一种语法好奇；语言中有没有一种方法可以在不将参数传递给 doIt 函数本身的情况下表达这一点?如果没有，这没什么大不了的，我已经很清楚我可以优雅地解决它的方法，因此无需提供替代解决方案.(我会接受不"作为答案，换句话说，如果这是事实的话.:-P)

This is pretty much purely a syntax curiosity; is there a way in the language to express this without passing arguments to the doIt function itself? If not, it's no big deal and I'm already well aware of ways I can gracefully work around it, so no need to provide alternative solutions. (I'll accept "no" as an answer, in other words, if that's the truth. :-P)