Get element by position (X and Y) in react-native - react-native

Question

On web, you can get any element by its position:
document.elementFromPoint(x, y)
Is there something similar in React Native? Can it be done with a native bridge (Java/Objective C)?
I would like to get an element on screen by position X and Y (Ignoring empty places and following transformation or styling calculations like padding, margin, borderRadius) then set it a native prop or dispatch events.
OBS: I'm not referring to onLayout property or any other declared in the component's construction/render, my idea is from any X and Y position get an element that corresponds to these coordinates then get a reference to it. A use case for example: Create a virtual cursor that dispatch click events on correct components, following margin/padding and ignoring pointerEvents none.
Image Example

Answer 1

You can use onLayout prop of React Native components.
You should check this link

Answer 2

Yes you can get your current elemnts position like in. 2 ways,
onLayout
suppose on click of that element you want to know that location
assign a ref to that
const newRef = useRef();
onButtonPress = () => {
newRef?.current?.measureInWindow( (fx, fy, width, height, px, py) => {
console.log('Component width is: ' + width)
console.log('Component height is: ' + height)
console.log('X offset to frame: ' + fx)
console.log('Y offset to frame: ' + fy)
console.log('X offset to page: ' + px)
console.log('Y offset to page: ' + py)
})
}
const onLayout=(event)=> {
const {x, y, height, width} = event.nativeEvent.layout;
}
return(
<TouchableOpacity ref={useRef} onLayout={onLayout} onPress={onButtonPress} >
</TouchableOpacity>
)