故意的int溢出 [英] On-purpose int overflow

问题描述

我正在使用哈希函数murmur2，该函数会向我返回uint64.

I'm using the hash function murmur2 which returns me an uint64.

然后我想将其存储在仅支持BIGINT(带符号的64位)的PostgreSQL中.

I want then to store it in PostgreSQL, which only support BIGINT (signed 64 bits).

因为我对数字本身不感兴趣，但对二进制值不感兴趣(因为我将其用作检测唯一性的ID(我的一组值约为1000个值，所以64位哈希对我来说足够了))希望通过只是"更改类型将其转换为int64.

As I'm not interested in the number itself, but just the binary value (as I use it as an id for detecting uniqueness (my set of values being of ~1000 values, a 64bit hash is enough for me) I would like to convert it into int64 by "just" changing the type.

如何以一种使编译器满意的方式进行操作?

How does one do that in a way that pleases the compiler?

推荐答案

您可以简单地使用 conversion :

i := uint64(0xffffffffffffffff)
i2 := int64(i)
fmt.Println(i, i2)

输出:

18446744073709551615 -1

将uint64转换为int64始终会成功:它不会仅更改类型的内存表示.如果您尝试将无类型的整数常量值转换为int64:

Converting uint64 to int64 always succeeds: it doesn't change the memory representation just the type. What may confuse you is if you try to convert an untyped integer constant value to int64:

i3 := int64(0xffffffffffffffff) // Compile time error!

这是一个编译时错误，因为常量0xffffffffffffffff(以任意精度表示)不适合int64，因为适合int64的最大值是0x7fffffffffffffff:

This is a compile time error as the constant value 0xffffffffffffffff (which is represented with arbitrary precision) does not fit into int64 because the max value that fits into int64 is 0x7fffffffffffffff:

constant 18446744073709551615 overflows int64

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