Reflection.Pointer（）返回的值不是值地址

Question

type BookInfo struct {
    Meta         *TableMeta
    ...
}

func (si *schemaInfo) getTabInfo(obj interface{}) (*tabInfo, error) {
    typ := reflect.TypeOf(obj)
    val := reflect.ValueOf(obj)
    if typ.Kind() != reflect.Ptr {
        return nil, errors.New("nborm.schemaInfo.getDBInfo() error: required a pointer")
    }
    meta := *(**TableMeta)(unsafe.Pointer(val.Pointer()))
    ...
 }

getTabInfo() works well, but I want to know why val.Pointer() return a value of **TableMeta? Why not a *TableMeta?The document of reflect says,

Pointer returns v's value as a uintptr. It returns uintptr instead of unsafe.Pointer so that code using reflect cannot obtain unsafe.Pointers without importing the unsafe package explicitly. It panics if v's Kind is not Chan, Func, Map, Ptr, Slice, or UnsafePointer.

In my mind:

info := &BookInfo{}
val := reflect.ValueOf(info)
ptr := val.Pointer()
meta := (*TableMeta)(unsafe.Pointer(val.Pointer()))

should work, but infact when I called val.Pointer(), the returned value is the pointer of *TableMeta(**TableMeta).

Answer 1

The value you have is a pointer to a BookInfo struct, it is of type *BookInfo. And the type of BookInfo.Meta field is also a pointer, it is of type *TableMeta, thus a *BookInfo can then be looked at as **TableMeta, hence the "double" pointer.

It's true that the struct pointer points to its first field, but don't build on it. It's fragile. If you add a field before it, it'll break badly (which will only happen at runtime, no compile time messages due to package unsafe).

So if the value is of type *BookInfo, simply obtain that out of the reflect.Value wrapper, then you can refer to its field like value.Meta, which will be of type *TableMeta. Avoid using package unsafe, especially if it's not needed.