基于注释，如果您不能将foo的签名更改为除原始函数指针之外的任何内容，那么您就必须这样做像这样的东西:

Based on the comments, if you cannot change the signature of foo to take anything but a raw function pointer... then you'll have to do something like this:

struct XFunc2Wrapper {
    static X* x;

    static void func2(int v) {
        x->func2(v);
    }
};

然后只要将XFunc2Wrapper::x设置为X，就执行foo(&XFunc2Wrapper::func2).它不必嵌套在结构中，它可以只是一些全局指针，但是嵌套有助于更好地建立代码背后的意图.

And then just do foo(&XFunc2Wrapper::func2) once you set XFunc2Wrapper::x to be your X. It doesn't have to be nested in a struct, it can just be some global pointer, but nesting helps establish the intent behind the code better.

但这绝对是(按照Obvlious上尉的)尝试执行foo(std::function<void(int)> )之后的不得已的手段.

But this should definitely be last resort after (as per Captain Obvlious) trying to do foo(std::function<void(int)> ).