I know this is an old question asked many times. but I cannot find a satisfactory answer for this, so I ask again. can someone explain what exactly happens in case of overflow of the whole and the lower stream? I heard about some lower order bytes that handle this, can anyone explain what this is?
thank!
You could imagine that when you only have 2 places that you count (so adding 1 each time)
00 01 10 11 100
"00". , "". 00. , , , , , . ( 11 00)
Mark peters : , , , . , "" .
Java ( , ).
:
System.out.println(Integer.MAX_VALUE + 1 == Integer.MIN_VALUE); System.out.println(Integer.MIN_VALUE - 1 == Integer.MAX_VALUE);
true.
, , 2- , , ( ) , .
, , , , Integer.MIN_VALUE - 1 Integer.MAX_VALUE.
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Another way to think about how java handles overflow / overclocking is to use an anology clock image. You can move it forward an hour at a time, but in the end the clock will start again. You can wind the clock back, but as soon as you go beyond the start, you are at the end again.