Java 如何检测 32 位 int 上的整数溢出？
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How can I detect integer overflow on 32 bits int?
提问by ashur
I know such topic was asked several times, but my question is about overflow on full 32 bits of int. For example:
我知道这样的话题被问过几次，但我的问题是关于 int 的完整 32 位溢出。例如：
11111111111111111111111111111111 +
00000000000000000000000000000001 =
00000000000000000000000000000000 //overflow!
I found topicwith similar question about this, however the algorithm is not perfect.
我发现了与此相关的类似问题的主题，但是该算法并不完美。
11111111111111111111111111111111 +
00000000000000000000000000000000 =
00000000000000000000000000000000 //overflow!
Is there any simple, fast, safer way to check this ?
有没有简单、快速、更安全的方法来检查这个？
采纳答案by Patryk Czarnik
Math.addExactthrows exception on overflow
Math.addExact溢出时抛出异常
Since Java 8 there is a set of methods in the Mathclass:
从 Java 8 开始，Math类中有一组方法：
…and versions for long as well.
......还有很长的版本。
Each of these methods throws ArithmeticExceptionif overflow happens. Otherwise they return the proper result if it fits within the range.
ArithmeticException如果发生溢出，这些方法中的每一个都会抛出。否则，如果它适合范围内，它们将返回正确的结果。
Example of addition:
添加示例：
int x = 2_000_000_000;
int y = 1_000_000_000;
try {
int result = Math.addExact(x, y);
System.out.println("The proper result is " + result);
} catch(ArithmeticException e) {
System.out.println("Sorry, " + e);
}
See this code run live at IdeOne.com.
Sorry, java.lang.ArithmeticException: integer overflow
抱歉，java.lang.ArithmeticException：整数溢出
回答by Tim B
long test = (long)x+y;
if (test > Integer.MAX_VALUE || test < Integer.MIN_VALUE)
// Overflow!
回答by Durandal
Overflow can be detected by a logical expression of the most significant bit of the two operands and the (truncated) result (I took the logical expression from the MC68030 manual):
溢出可以通过两个操作数的最高有效位的逻辑表达式和（截断的）结果来检测（我从MC68030手册中获取了逻辑表达式）：
/**
* Add two int's with overflow detection (r = s + d)
*/
public static int add(int s, int d) throws ArithmeticException {
int r = s + d;
if (((s & d & ~r) | (~s & ~d & r)) < 0)
throw new ArithmeticException("int overflow add(" + s + ", " + d + ")");
return r;
}
回答by GOTO 0
The most intuitive method I can think of: calculate the sum (or difference) as a long, then convert that sum to an intand see if its value has changed.
我能想到的最直观的方法是：将总和（或差）计算为 a long，然后将该总和转换为 anint并查看其值是否已更改。
long longSum = (long) a + b;
int sum = (int) longSum;
if (sum == longSum) {
// sum contains the correct result
} else {
// overflow/underflow
}
Remember that on modern 64 bit processors, working with longs is no less efficient than working with ints (the opposite may be true). So if you have a choice between checking for overflows or using longs, go for the latter.
请记住，在现代 64 位处理器上，使用longs 并不比使用ints效率低（反之亦然）。因此，如果您可以在检查溢出或使用longs之间做出选择，请选择后者。
回答by Vusal
Try this way:
试试这个方法：
boolean isOverflow(int left, int right) {
return right > 0
? Integer.MAX_VALUE - right < left
: Integer.MIN_VALUE - right > left;
}
From: https://wiki.sei.cmu.edu/confluence/display/java/NUM00-J.+Detect+or+prevent+integer+overflow
来自：https: //wiki.sei.cmu.edu/confluence/display/java/NUM00-J.+Detect+or+prevent+integer+overflow
