react native onChangeText and onChange will give input value missing 1 character - react-native

Question

I was wondering why there's 1 missing character every time I type on text input using onChangeText and onChange function simultaneously.
Here's my code:
import React, {useState} from "react";
import {Text} from "galio-framework";
export default function UpcomingSurvey ({ navigation }) {
const [timer, setTimer] = useState('')
const [saving, setSaving] = useState(false);
const [edited, setEdited] = useState(false);
const [inputValue, setInputValue] = useState('')
const inputChanged = () => {
setSaving(true)
clearTimeout(timer)
const newTimer = setTimeout(() => {
console.log('inputValue: ', inputValue)
}, 2000)
setTimer(newTimer)
}
function renderUpcomingSurvey () {
<Block style={{ paddingHorizontal: theme.SIZES.BASE }}>
<>
{saving &&
<Block style={{flex: 1}}>
<Text muted size={11} style={{ textAlign: 'right'}}>Saving</Text>
</Block>
}
{(edited && !saving) &&
<Block style={{flex: 1}}>
<Text muted size={11} style={{textAlign: 'right', color: '#36d79a'}}>saved</Text>
</Block>
}
</>
{!isLoading ? (
<Input
editable={radioCustom}
onChangeText={setInputValue}
onChange={inputChanged}
value={inputValue}
right placeholder="Type your custom question here."
iconContent={<Block />}
/>) : <Text></Text>
}
<Text muted size={12}>Free text | Your employees will see that this is a custom question</Text>
</Block>
When I try to console.log('inputValue: ', inputValue) inside inputChanged function will give missing one character. Example when I try to input Text this will give value in console Tex which missing character t.
Maybe there's a delay on capturing text value input after the onChange function triggers
What's the real culprit of these code?

Answer 1

It is probably the async nature of the setState function
Try to log the input value like this:
useEffect(() => {
console.log(inputValue)
}, [inputValue]) //Only run when inputValue is set

Answer 2

Solve it by assigning the Input defaultValue={inputValue} instead of value={inputValue}