如何将十六进制长字符串解析为uint [英] How to parse long hexadecimal string into uint

问题描述

我要提取十六进制字符串形式的数据，我需要将其转换为十进制表示法，截断18个小数位，然后在JSON中使用.

I'm pulling in data that is in long hexadecimal string form which I need to convert into decimal notation, truncate 18 decimal places, and then serve up in JSON.

例如，我可能具有十六进制字符串:

For example I may have the hex string:

"0x00000000000000000000000000000000000000000000d3c21bcecceda1000000"

起初我尝试使用ParseUint()，但是由于它支持的最高级别是int64，所以我的电话号码最终太大了.

At first I was attempting to use ParseUint(), however since the highest it supports is int64, my number ends up being way too big.

此示例在转换和截断后的结果为10 ^ 6. 但是，在某些情况下，此数字最多可以为10 ^ 12(意味着截断为10 ^ 30！).

This example after conversion and truncation results in 10^6. However there are instances where this number can be up to 10^12 (meaning pre truncation 10^30!).

攻击此问题的最佳策略是什么?

What is the best strategy to attack this?

推荐答案

使用数学/大用于处理大于64位的数字.

Use math/big for working with numbers larger than 64 bits.

来自 Int.SetString 示例:

s := "d3c21bcecceda1000000"
i := new(big.Int)
i.SetString(s, 16)
fmt.Println(i)

https://play.golang.org/p/vf31ce93vA

math/big类型还支持encoding.TextMarshaler和fmt.Scanner接口.

例如

i := new(big.Int)
fmt.Sscan("0x000000d3c21bcecceda1000000", i)

或

i := new(big.Int)
fmt.Sscanf("0x000000d3c21bcecceda1000000", "0x%x", i)

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