I need help with code and explains how it works :)

Hello everyone I need some help on how to make an active program.
I have an encoder, Lcd, and DHT.
the program does not work because there is a return
how to bypass return

#include <Scheduler.h>
#include <DHT.h>
#include <LiquidCrystal.h>
#define DHT11_PIN 6

const int rs = 12, en = 11, d4 = 5, d5 = 4, d6 = 3, d7 = 2;
LiquidCrystal lcd(rs, en, d4, d5, d6, d7);

const int clkPin = 7; //the clk attach to pin 2
const int dtPin = 8; //the dt pin attach to pin 3
const int swPin = 9 ; //the sw pin attach to pin 4

int encoderVal = 0;
dht DHT;

void setup()
{
  Serial.begin(9600);

  pinMode(clkPin, INPUT);
  pinMode(dtPin, INPUT);
  pinMode(swPin, INPUT);
  digitalWrite(swPin, HIGH);

  Scheduler.startLoop(loop2);

  lcd.begin(16, 2);
}
void loop()
{
  lcd.setCursor(3, 0);
  int change = getEncoderTurn();
  encoderVal = encoderVal + change;
  if (digitalRead(swPin) == LOW)
  {
    encoderVal = 0;
  }
  Serial.println(encoderVal);
  lcd.setCursor(3, 0);
  lcd.print(encoderVal);
  lcd.print(" ");
}
void loop2();
{
  int chk = DHT.read11(DHT11_PIN);
  lcd.setCursor(0, 1);
  lcd.print("T");
  lcd.setCursor(1, 1);
  lcd.println(DHT.temperature);
  lcd.setCursor(9, 1);
  lcd.print("H");
  lcd.setCursor(10, 1);
  lcd.print(DHT.humidity);
  delay(1500);
  yield();
}
int getEncoderTurn(void)
{
  static int oldA = HIGH; //set the oldA as HIGH
  static int oldB = HIGH; //set the oldB as HIGH
  int result = 0;
  int newA = digitalRead(clkPin);//read the value of clkPin to newA
  int newB = digitalRead(dtPin);//read the value of dtPin to newB
  if (newA != oldA || newB != oldB)
  {
    // something has changed
    if (oldA == HIGH && newA == LOW)
    {
      result = (oldB * 2 - 1);
    }
  }
  oldA = newA;
  oldB = newB;
  return result;
}

The command "return" only appears in line 75, and seems perfectly normal. Why do you think it is causing a problem?

You need to tell us exactly what your program actually does and what you want it to do that is different. The words "does not work" do not provide any useful information with which to help you.

...R