C语言 如何在没有未定义行为的情况下检查 C 中的有符号整数溢出？
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How to check for signed integer overflow in C without undefined behaviour?
提问by Ganker
There's (1):
有（1）：
// assume x,y are non-negative
if(x > max - y) error;
And (2):
和（2）：
// assume x,y are non-negative
int sum = x + y;
if(sum < x || sum < y) error;
Whichs is preferred or is there a better way.
这是首选还是有更好的方法。
回答by caf
Integer overflow is the canonical example of "undefined behaviour" in C (noting that operations on unsigned integers never overflow, they are defined to wrap-around instead). This means that once you've executed x + y, if it overflowed, you're already hosed. It's too late to do any checking - your program could have crashed already. Think of it like checking for division by zero - if you wait until after the division has been executed to check, it's already too late.
整数溢出是 C 中“未定义行为”的典型示例（注意对无符号整数的操作永远不会溢出，它们被定义为环绕）。这意味着一旦你执行了x + y，如果它溢出，你已经被冲洗掉了。现在进行任何检查都为时已晚 - 您的程序可能已经崩溃了。把它想象成检查除以零 - 如果你等到除法执行完后再检查，已经太晚了。
So this implies that method (1) is the only correct way to do it. For max, you can use INT_MAXfrom <limits.h>.
所以这意味着方法 (1) 是唯一正确的方法。对于max，您可以使用INT_MAXfrom <limits.h>。
If xand/or ycan be negative, then things are harder - you need to do the test in such a way that the test itself can't cause overflow.
如果x和/或y可能是负数，那么事情就更难了 - 您需要以一种测试本身不会导致溢出的方式进行测试。
if ((y > 0 && x > INT_MAX - y) ||
(y < 0 && x < INT_MIN - y))
{
/* Oh no, overflow */
}
else
{
sum = x + y;
}
回答by Chris Dodd
You can really only check for overflow with unsignedintegers and arithmatic:
你真的只能用unsigned整数和算术检查溢出：
unsigned a,b,c;
a = b + c;
if (a < b) {
/* overflow */
}
The behavior of overflow with signed integers is undefined in C, but on most machines you can use
有符号整数溢出的行为在 C 中是未定义的，但在大多数机器上你可以使用
int a,b,c;
a = b + c;
if (c < 0 ? a > b : a < b) {
/* overflow */
}
This won't work on machines that use any kind of saturating arithmetic
这不适用于使用任何类型的饱和算法的机器
回答by clahey
You only have to check one of them. If x + y overflows, it will be less than both x and y. Hence:
您只需要检查其中之一。如果 x + y 溢出，它将小于 x 和 y。因此：
int sum = x + y;
if (sum < x) error;
should be sufficient.
应该足够了。
The following site has a bunch of stuff about integer overflow:
以下站点有一堆关于整数溢出的内容：
If you want to handle negative numbers, it can be expanded:
如果要处理负数，可以展开：
int sum = x + y;
if (y >= 0) {
if (sum < x) error;
} else {
if (sum > x) error;
}
