Question : Java Swing logical operator

I have some codes extracted from DefaultButtonModel.java. I am wondering what does following operation mean:

stateMask |= ENABLED;
stateMask &= ~PRESSED;

stateMask is int, ENABLED is final int (1 << 3), PRESSED is int(1<<2).

public final static int PRESSED = 1 << 2;
public final static int ENABLED = 1 << 3;

Does stateMask |= ENABLED mean: 1) stateMask OR ENABLED, then assign it back to stateMask?
Does stateMask &= ~PRESSED mean: 1) neg PRESS first (0x0 to 0x1, 0x1 to 0x0), then AND with stateMask, result is assigned to stateMask?

         
           
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          public void setEnabled(boolean b) {
        if(isEnabled() == b) {
            return;
        }
            
        if (b) {
            stateMask |= ENABLED;
        } else {
            stateMask &= ~ENABLED;
	    // unarm and unpress, just in case
            stateMask &= ~ARMED;
            stateMask &= ~PRESSED;
        }

            
        fireStateChanged();
    }
        
    /**
     * {@inheritDoc}
     */
    public void setSelected(boolean b) {
        if (this.isSelected() == b) {
            return;
        }

        if (b) {
            stateMask |= SELECTED;
        } else {
            stateMask &= ~SELECTED;
        }

        fireItemStateChanged(
                new ItemEvent(this,
                              ItemEvent.ITEM_STATE_CHANGED,
                              this,
                              b ?  ItemEvent.SELECTED : ItemEvent.DESELECTED));
        
        fireStateChanged();
        
    }

           
         

Answer : Java Swing logical operator

Hi learningunix,

the line checks if the left-most byte of 'num' is '1'.

'&num' is a pointer to the memory address where the first byte of 'num' resides. The '(char)' casts this pointer '&num' (which is a pointer to int) to a pointer to char. Since char is a one byte data type accessing that 'pointer to char' with '' accesses the first byte of the int. In little endian this byte has to be '1' for and 'int' which is '1' - in big endian the first byte would be '0' since the least significant byte is the most right one ...

Hope that helps,

ZOPPO