what happens to the last return *this c++?

Question

  Time &Time::setHour( int h ) 
  {
     hour = ( h >= 0 && h < 24 ) ? h : 0; 
     return *this; 
  } 


  Time &Time::setMinute( int m ) 
  {
     minute = ( m >= 0 && m < 60 ) ? m : 0; 
     return *this;  
  } 


  Time &Time::setSecond( int s ) 
  {
     second = ( s >= 0 && s < 60 ) ? s : 0; 
    return *this; 
   }

int main()
{
    Time t;     
    t.setHour( 18 ).setMinute( 30 ).setSecond( 22 );
    return 0;
}

I understand the cascaded member function call, but I don't understand how t.setHour( 18 ).setMinute( 30 ).setSecond( 22 ); is left hanging, doesn't it have to be assigned to something since it still returns *this after it's done cascading? Why is it ok to leave it like that?

Answer 1

The value of the expression which is used as a statement is discarded. It's the same thing that happens when you write

scanf("%d", &i);

You did know that scanf has a return value, right? And so does ++i in

for( int i = 0; i < 10; ++i )

And even

x = 5;

is throwing away the result of the expression.

There are lots of expressions and functions with return values that are only marginally useful. If you want to write the code in a way that makes it clear the return value is ignored intentionally and not by accident, cast to void:

(void) scanf("%d", &i);

This should silence compiler warnings about ignored return values.

These variations are also valid but a bit sillier:

(void)++i;
(void)(x = 5);