在 Kotlin 中编写一个等于 Int.MIN_VALUE 的十六进制整数文字

发布于07月04日

how does one write a hexadecimal integer literal that is equal to Int.MIN_VALUE (which is -2147483648 in decimal) in Kotlin?

AFAIK, an Int is 4 bytes...and sometimes it seems like 2's complement is used to represent integers...but I'm not sure. I've tried the following hex literals to help myself understand the system:

0xFFFFFFFF但这是Long而不是Int
0xFFFFFFFF.toInt() which is -1
-0xFFFFFFFF.toInt()等于1
0x7FFFFFFF是2147483647是Int.MAX_VALUE
-0x7FFFFFFF which is -2147483647 which is Int.MIN_VALUE+1
0xFFFFFFF which is 268435455 in decimal
0x0FFFFFFF which is also 268435455 in decimal

But I can't figure out what hexadecimal integer literal can be used to represent Int.MIN_VALUE.

我希望答案不会让我觉得自己很愚蠢...

推荐答案

Int represents a 32-bit signed integer. 32 bits means 8 hex digits:

___7 F F F F F F F

0111 1111 1111 1111 1111 1111 1111 1111

可以看到，最左边的位是0，因此这是32位表示中的正积分.根据2's complement的定义和示例，最小32位负值最左边的位是1，后面是0:

1000 0000 0000 0000 0000 0000 0000 0000个

___8 0 0 0 0 0 0 0

一共是0x80000000美元.

In Kotlin you need to prepend the - sign to denote negative Int which is not true in Java. Consider following example

println(0x7FFFFFFF) // -> prints 2147483647 (Integer.MAX_VALUE)
println(-0x80000000) // -> prints -2147483648 (Integer.MIN_VALUE)
println(0x80000000) // -> prints 2147483648 (does not fit into Int)

这与Java不同:

System.out.println(0x7FFFFFFF); // -> prints 2147483647 (Integer.MAX_VALUE)
System.out.println(-0x80000000); // -> prints -2147483648 (Integer.MIN_VALUE)
System.out.println(0x80000000); // -> prints -2147483648 (Integer.MIN_VALUE)

This is in line with Kotlin spec although the overflow behavior of hexadecimal literals is yet to be defined.

进一步阅读: