Consider the following code:

long y = 9223372036854775807L;
y = y + 1;
System.out.println(y);

What number does the program print when compiled and run? Hint: think about the max value you just looked up in your previous answer.

Question

Consider the following code:

long y = 9223372036854775807L;
y = y + 1;
System.out.println(y);

What number does the program print when compiled and run? Hint: think about the max value you just looked up in your previous answer.

BlackTom AI · Accepted Answer

First, recall the behavior of 64-bit signed integers in Java. A long is a 64-bit two's complement integer, with a range from -2^63 to 2^63 - 1, i.e., from -9223372036854775808 to 9223372036854775807. This means that when you add 1 to the maximum value, the value wraps around to the minimum value due to overflow.

Option 1: -9223372036854775808. This matches the expected result of overflowing Long.MAX_VALUE by 1, yielding Long.MIN_VALUE. So this option aligns with how Java handles overflow for long arithmetic.

Option 2: -1. This would imply an overflow from a small negative to another negative, or some other erroneous state, but it does not reflect the standard 64-bit overflow wraparound from the exact maximum value. This choice misinterprets the wrap behavior.

Option 3: 0. Getting 0 would require a different operation (for example, adding 1 to  -1 with a specific sequence that causes modulo behavior in a loop), but directly adding 1 to Long.MAX_VALUE cannot produce 0 in a single overflow step.

Option 4: This program crashes and raises a longOverflow exception. In Java, integer overflow is defined by wraparound semantics for primitive integral types; there is no exception thrown for overflow. Therefore, a runtime crash or a specific longOverflow exception does not occur here.

Option 5: 9223372036854775808. This value is one greater than Long.MAX_VALUE and cannot be represented in a long. Instead, the overflow wraps to Long.MIN_VALUE, not this out-of-range value.

In summary, understanding the fixed bounds of long and the wraparound behavior on overflow explains why the value after y = y + 1; is Long.MIN_VALUE, which corresponds to -9223372036854775808.

Consider the following code: long y = 9223372036854775807L; y = y + 1; System.out.println(y); What number does the program print when compiled and run? Hint: think about the max value you just looked up in your previous answer.单项选择题

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