Golang vs Python-十六进制字符串到整数

Question

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?

Answer 1

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))