Parsing Long Hexadecimal Strings into uint with High Precision
Converting long hexadecimal strings into decimal notation with high precision is a common task in various programming scenarios. However, the default ParseInt()/ParseUint() functions in Go have limitations when working with numbers that exceed 64-bit precision.
To address this issue, the recommended approach is to utilize the math/big package in Go. This package provides support for working with integers of arbitrary size, enabling you to handle numbers that are far larger than what the standard int64 type can accommodate.
Using math/big for High-Precision Parsing
The math/big package provides a big.Int type that represents integers with arbitrary precision. To convert a hexadecimal string into a big.Int, you can use the following steps:
Example with math/big
Here's an example of how to use math/big to parse a long hexadecimal string into a big.Int:
import ( "fmt" "math/big" ) func main() { s := "0x00000000000000000000000000000000000000000000d3c21bcecceda1000000" i := new(big.Int) i.SetString(s, 16) fmt.Println(i) }
This code converts the provided hexadecimal string into a big.Int and prints the result.
Additional Features of math/big
The math/big package also provides support for the encoding.TextMarshaler and fmt.Scanner interfaces. This allows you to easily read and write big.Int values from and to text-based sources.
For instance, you can use the following syntax to parse a hexadecimal string using fmt.Sscan():
var i big.Int fmt.Sscan("0x000000d3c21bcecceda1000000", &i)
Alternatively, you can use fmt.Sscanf() to parse the hexadecimal string directly into a big.Int:
var i big.Int fmt.Sscanf("0x000000d3c21bcecceda1000000", "0x%x", &i)
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