Question

我正在使用Golang的RSA加密库。以下是加密消息的功能：

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

random参数用作熵的来源，以确保两次对同一消息进行加密不会产生相同的密文。此参数允许使用Linux函数，例如getrandom（2）（如果可用）或/ dev / urandom，否则可以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（即在EncryptOAEP函数外部），然后将其插入其中。这个库有可能吗？如果不是这样，我可以使用什么Golang库来实现此功能？我希望我足够清楚，感谢您的帮助！

Answer 1

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

这是Go crypto/rsa包中的示例。我用crypto/rand io.Reader取代了math/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()
}

游乐场：https://play.golang.org/p/KtqUuDC2Tai

输出：

Ciphertext: 47516ff9863897f8f5344d07bca363515856ace56fed5451ede414eb6771eff357f09ca67f07d0d918b6e7107975f83afc4eebc932a7d9ba3f48c3f70399af88917c852e46a3d43eaa9bc563c7780bf7cf090c0b5e1690ac77af9e7c8cc6ec0d2fb729eed3fb45c824754bfd898e02eee8dae88b4fb3277f6970dbb0b929da7c14d245e51ed3a3e63127efc3e4c438adcfce78bbc1117a134f5a9dd62f9ebc14f125561c959ce6963a2b8605687ab568ec570295bb8f25477c7bbcb7303d96e888a0150e23ea0dd4880fefe40d86567518784d7f54c594f2c0e576a14c210201bf9d75fbd96b70fdd793c28ffecaa5c036b2949dba6e66004ec2f0e676095f1c

如何自己产生熵？ -RSA Golang

1 个答案: