How to return a value from a void function and use it inside the void loop?

Using Arduino Programming Questions
4

@KeithR
Thanks, I removed the return print() from the If statement inside the void print() function.

@Delta_G

Thank you for your suggestions.

I followed your examples

int intPressure = 12;  
int tweet = 0;
int val = 0;

void setup() {
}


void print() {
  
  if(intPressure != 0)
  {  
    Serial.println("HIGH");
    tweet = HIGH;
  }
  else 
  {
    Serial.println("LOW");
    tweet = LOW;
  }
   return tweet;
}

void loop() {

  int tweetStatus = print();
  
  val = digitalRead(LED_BUILTIN);
  digitalWrite(LED_BUILTIN, tweetStatus);
  Serial.println("---");
  Serial.println(val);
  Serial.println("---");


}

and got 2 error messages :

Error A:
In function 'void print()':
test_bed:21: error: 
return-statement with a value, in function returning 'void' [-fpermissive]
    return tweet; 
           ^

Error B: 
In function 'void loop()':
test_bed:26: error: 
void value not ignored as it ought to be
   int tweetStatus = print(); 
                           ^
exit status 1
return-statement with a value, in function returning 'void' [-fpermissive]

Despite the explanations regarding the error messages below, I don't understand them quite, yet:

Error A
It means a function is declared to return nothing (void) but you are assigning the return code to a variable.

source: void value not ignored as it ought to b - C++ Forum

Error B
The error means that in your function:
void* foo(...);
You have a statement:
return;
But the compiler expects a value to be provided:
return myVoidPtr;

source: c++ - error: return-statement with no value, in function returning ‘void*’ [-fpermissive] - Stack Overflow

Are the error messages saying that the return statement return tweet returns something that is not there?