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How to Handle Hexadecimal Strings Larger Than Go's Built-in Integer Types?

Patricia Arquette
Release: 2024-12-15 18:08:09
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How to Handle Hexadecimal Strings Larger Than Go's Built-in Integer Types?

How to Parse Oversized Hexadecimal Strings in Go

Converting hexadecimal strings into decimal numbers can be a challenge when dealing with numbers that exceed the capacity of built-in integer types. Golang provides a powerful solution with its math/big package.

Converting Hexadecimal Strings to Big Integers

To convert a hexadecimal string into a big.Int, use the SetIntString method.

package main

import "math/big"

func main() {
    // Example hexadecimal string
    hexString := "0x00000000000000000000000000000000000000000000d3c21bcecceda1000000"

    // Create a new big.Int
    i := new(big.Int)

    // Convert the hexadecimal string to a big.Int
    i.SetString(hexString, 16)

    // Print the converted number
    println(i)
}
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Truncating Decimal Representation

After converting the hexadecimal string to a big.Int, you can truncate the decimal representation by using the Truncate method. For example:

func Truncate(x *big.Int, prec uint) {
    f := new(big.Float).SetInt(x)

    f.SetPrec(int64(prec))
    f.Int(x)
}
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This function sets the precision of the big.Float to the desired number of decimal places before converting it back to a big.Int.

Usage Examples

You can use the fmt package to parse and format big.Int values:

package main

import "fmt"

func main() {
    hexString := "0x000000d3c21bcecceda1000000"

    // Create a new big.Int
    i := new(big.Int)

    // Parse the hexadecimal string
    fmt.Sscan(hexString, i)

    // Alternatively, you can use fmt.Sscanf
    // fmt.Sscanf(hexString, "0x%x", i)

    // Truncate to 18 decimal places
    Truncate(i, 18)

    // Marshal to JSON
    jsonBytes, _ := i.MarshalJSON()
    fmt.Println(string(jsonBytes))
}
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