为什么以下程序会发生溢出？-Whymayanoverflowoccurinthefollowingprogram?

13  

An overflow may occur because the range of integer representation in two's complement is not symmetric: the magnitude of the smallest negative number that can be represented is the magnitude of the highest positive number that can be represented, plus one. For example, on a 32-bit system the values are -2,147,483,648 and 2,147,483,647. That's why negating -2,147,483,648 would result in an overflow: the result of negation, a positive value 2,147,483,648, cannot be represented in an int of the same size.

可能会发生溢出,因为二进制补码中的整数表示范围不对称:可以表示的最小负数的大小是可以表示的最高正数的大小加一。例如,在32位系统上,值为-2,147,483,648和2,147,483,647。这就是为什么否定-2,147,483,648会导致溢出:否定的结果,正值2,147,483,648,不能用相同大小的int表示。

Note that the inverse of this problem is not true: negating a positive number would not result in an overflow:

请注意,此问题的反之亦然:否定正数不会导致溢出:

if (i > 0) { i = -i; } // No overflow here

1  

Your value of "i", in the stack, is undefined when main starts. The start-up code that runs before main() is called can leave anything there.

堆栈中的“i”值在主启动时未定义。调用main()之前运行的启动代码可以在那里留下任何东西。

Addig to that what Kashif said, negative integers can go one value lower than non-negative integers as negatives don't need to leave room for zero. A "1" in the sign bit, with all remaining bits zero, causes an overflow when sign reversed.

加上Kashif所说的,负整数可以比非负整数低一个值,因为负数不需要留下零空间。符号位中的“1”,所有剩余的位为零,当符号反转时会导致溢出。

In 16 bits: -0x8000 == ~0x8000 + 1 == 0x7FFF + 1 == 0x8000
// "-" negative value == invert + 1 == inverted + 1 == final is the same

在16位:-0x8000 == ~0x8000 + 1 == 0x7FFF + 1 == 0x8000 //“ - ”负值==反转+ 1 ==倒置+ 1 ==最终是相同的

The likely hood of this value is low, but present. It will not happen unless the stack just happens to contain the offending number.

这个值的可能性很低,但存在。除非堆栈恰好包含有问题的数字,否则不会发生。

0  

The cause for an integer overflow is when an arithmetic operation attempts to create a numeric value that is outside of the range that can be represented with a given number of bits, either larger than the maximum or lower than the minimum representable value.

整数溢出的原因是算术运算尝试创建一个数值,该数值超出了可以用给定位数表示的范围,该位数大于最大值或小于最小可表示值。

• Well, in your case the variable i is not initialized. So what will happen here is that the memory space which is assigned to variable i of integer type will be containing some garbage value.

那么,在你的情况下,变量i没有初始化。那么这里会发生的是,分配给整数类型的变量i的内存空间将包含一些垃圾值。

• If by any chance that memory address contains the maximum possible integer value which is -2^31(-2,147,483,648) then negating this value will result in integer overflow.

如果内存地址有可能包含最大可能的整数值-2 ^ 31(-2,147,483,648),则否定该值将导致整数溢出。

I hope this helps.

我希望这有帮助。