go - 如何自己产生熵？ -RSA 戈朗

Question

我正在使用 Golang 的 RSA 加密库。下面是加密消息的函数:

func EncryptOAEP(hash hash.Hash, random io.Reader, pub *PublicKey, msg []byte, label []byte) ([]byte, error)

随机参数用作熵源，以确保对同一消息加密两次不会产生相同的密文。此参数允许使用 linux 函数，如 getrandom(2)(如果可用)或/dev/urandom(否则通过访问从设备和其他来源收集的环境噪声来充当随机数生成器)。我不希望 EncryptOAEP 函数使用操作系统的函数(getrandom(2) 或/dev/urandom)来生成随机数。我需要自己生成一个随机数(也就是在EncryptOAEP函数之外)，然后插入进去。这个图书馆有可能吗？如果不是，我可以使用什么 Golang 的库来实现此行为？我希望我说得足够清楚，我们将不胜感激，谢谢!

Answer 1

通过设计，这是可能的，而且很容易做到。提供任何 io.Reader 作为 random 参数。

这是来自 Go crypto/rsa 包的示例。我用 math/rand io.Reader 替换了 crypto/rand io.Reader 。

package main

import (
    "crypto/rsa"
    "crypto/sha256"
    "fmt"
    "math/big"
    "math/rand"
    "os"
    "time"
)

func main() {
    secretMessage := []byte("send reinforcements, we're going to advance")
    label := []byte("orders")

    // crypto/rand.Reader is a good source of entropy for randomizing the
    // encryption function.
    // rng := rand.Reader
    // However, you can use any io.Reader.
    // For example, math/rand.Reader
    rng := rand.New(rand.NewSource(time.Now().UnixNano()))

    ciphertext, err := rsa.EncryptOAEP(sha256.New(), rng, &test2048Key.PublicKey, secretMessage, label)
    if err != nil {
        fmt.Fprintf(os.Stderr, "Error from encryption: %s\n", err)
        return
    }

    // Since encryption is a randomized function, ciphertext will be
    // different each time.
    fmt.Printf("Ciphertext: %x\n", ciphertext)
}

func fromBase10(base10 string) *big.Int {
    i, ok := new(big.Int).SetString(base10, 10)
    if !ok {
        panic("bad number: " + base10)
    }
    return i
}

var test2048Key *rsa.PrivateKey

func init() {
    test2048Key = &rsa.PrivateKey{
        PublicKey: rsa.PublicKey{
            N: fromBase10("14314132931241006650998084889274020608918049032671858325988396851334124245188214251956198731333464217832226406088020736932173064754214329009979944037640912127943488972644697423190955557435910767690712778463524983667852819010259499695177313115447116110358524558307947613422897787329221478860907963827160223559690523660574329011927531289655711860504630573766609239332569210831325633840174683944553667352219670930408593321661375473885147973879086994006440025257225431977751512374815915392249179976902953721486040787792801849818254465486633791826766873076617116727073077821584676715609985777563958286637185868165868520557"),
            E: 3,
        },
        D: fromBase10("9542755287494004433998723259516013739278699355114572217325597900889416163458809501304132487555642811888150937392013824621448709836142886006653296025093941418628992648429798282127303704957273845127141852309016655778568546006839666463451542076964744073572349705538631742281931858219480985907271975884773482372966847639853897890615456605598071088189838676728836833012254065983259638538107719766738032720239892094196108713378822882383694456030043492571063441943847195939549773271694647657549658603365629458610273821292232646334717612674519997533901052790334279661754176490593041941863932308687197618671528035670452762731"),
        Primes: []*big.Int{
            fromBase10("130903255182996722426771613606077755295583329135067340152947172868415809027537376306193179624298874215608270802054347609836776473930072411958753044562214537013874103802006369634761074377213995983876788718033850153719421695468704276694983032644416930879093914927146648402139231293035971427838068945045019075433"),
            fromBase10("109348945610485453577574767652527472924289229538286649661240938988020367005475727988253438647560958573506159449538793540472829815903949343191091817779240101054552748665267574271163617694640513549693841337820602726596756351006149518830932261246698766355347898158548465400674856021497190430791824869615170301029"),
        },
    }
    test2048Key.Precompute()
}

Playground :https://play.golang.org/p/KtqUuDC2Tai

输出:

Ciphertext: 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