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Exceding the maximum value of a long doesnt throw an exception, instead it cicles back. If you do this: ... Find the answer to your question by asking. Ask question ... See similar questions with these tags. ... New site design and philosophy for Stack Overflow: Starting February 24, 2026... 278 How does Java ... More on stackoverflow.com

First off Integer.MIN_VALUE is -(231) and not 231 Integer.MAX_VALUE = 231 -1 = 2147483647. If we do the addition in binary we get: 1111111111111111111111111111111 +1111111111111111111111111111111 11111111111111111111111111111110 Which is -2 in decimal. As far as I remember the 32nd bit being 1 indicates that the number is negative. The binary representation of 231 - 1 takes 31 bits, the 32nd bit being 0. The operation of MAX_VALUE + MAX_VALUE overflows and makes the 32nd bit 1. To look at it another way.... Integer.MAX_VALUE + 1 = Integer.MIN_VALUE so Integer.MAX_VALUE + Integer.MAX_VALUE + 1 = Integer.MIN_VALUE + Integer.MAX_VALUE which implies Integer.MAX_VALUE + Integer.MAX_VALUE = Integer.MIN_VALUE + Integer.MAX_VALUE - 1 = -(231) + (231 -1) -1 = -(231) + (231) -1 -1 = -2 More on reddit.com

You want to test if the expression A * B > max is true, but you can't directly evaluate the term A * B because it might overflow, so adjust the inequality, perhaps A > max / B. (There could be rounding errors to account for.) Alternatively, you could compute A * B using a long. More on reddit.com