write a hexadecimal integer literal equal to Int.MIN_VALUE in Kotlin

问题

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 but this is a Long, not an Int
0xFFFFFFFF.toInt() which is -1
-0xFFFFFFFF.toInt() which is 1
0x7FFFFFFF which is 2147483647 which is 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.

I hope the answer doesn't make me feel stupid...

回答1:

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

As you can see the left-most bit is 0 thus this is a positive integral in a 32 bit representation. By 2's complement definition and example the minimal 32-bit negative value will have 1 at left-most bit followed by 0:

1000 0000 0000 0000 0000 0000 0000 0000

___8 0 0 0 0 0 0 0

that is 0x80000000.

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)

It's not the same as in Java:

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.