2019-07-05 19:05
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Golang vs Python-十六进制字符串到整数

I have a hex string:

n = "0xd458985bc81e284609dd69267c73b8464e1795d5b91ce6ed8871ecbc5b6ec4d1"  

that I can convert to an int in python by using:

mynum = int(n,16)

and I get the long number: 96046857981227695367604088053507399752198003710848334588478940192231467697361

Now how would I do this in Golang?

图片转代码服务由CSDN问答提供 功能建议


  n =“ 0xd458985bc81e284609dd69267c73b8464e1795d5b91ce6ed8871ecbc5b6ec4d1  “ 


  mynum = int(n,  16)

,我得到一个长号: 96046857981227627695367604088053507399752198003710848334588478940192231467697361

现在我该怎么办 这个在Golang吗?

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1条回答 默认 最新

  • duanbing8817 2019-07-06 04:58

    This is a good question (though similar to another that Flimzy found). The main problem is that the built-in integers only go up to 64 bits in Go but you can use the math/big package.

    fmt.Println(new(big.Int).SetString("0xd458985bc81e284609dd69267c73b8464e1795d5b91ce6ed8871ecbc5b6ec4d1", 0))

    If you don't have a leading 0x then use 16 as the base:

    fmt.Println(new(big.Int).SetString("d458985bc81e284609dd69267c73b8464e1795d5b91ce6ed8871ecbc5b6ec4d1", 16))
    解决 无用
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