如何使用go将十六进制转换为int [重复]
问题描述:
This question already has an answer here:
I converting hex value to decimal using go using below mentioned code
numberStr := strings.Replace(s, "0x", "", -1)
numberStr = strings.Replace(numberStr, "0X", "", -1)
n, err := strconv.ParseUint(numberStr, 16, 64)
if err != nil {
panic(err)
}
return n
Hexa value is : 0x340aad21b3b700000
but thrown error : strconv.ParseUint: parsing "340aad21b3b700000": value out of range
Can you suggest any alternate solution.
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此问题已经存在 在这里有答案 p>
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如何将十六进制长字符串解析为uint
1个答案
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我使用下面提到的代码使用go将十六进制值转换为十进制 p>
numberStr:=字符串。替换(s,“ 0x”,“”,-1) numberStr =字符串.Replace(numberStr,“ 0X”,“”,-1)\ nn,err:= strconv.ParseUint(numberStr,16,64) 如果err!= nil { panic(err) } 返回n code> pre>
十六进制值是:
0x340aad21b3b700000 code> p>
但抛出错误:
strconv.ParseUint:解析“ 340aad21b3b700000”:值超出范围 p>
您能提出任何其他解决方案吗? p> div>
答
Maximum value for uint64 is 0xFFFF FFFF FFFF FFFF, thus to use overflowing value you have to resort to package math/big of the standard library:
import "math/big"
...
n := new(big.Int)
n.SetString(numberStr, 16)
Continue with the package documentation.