如何将 JWK 中的私钥加载到 openSSL 中? [英] How to load a private key from a JWK into openSSL?
本文介绍了如何将 JWK 中的私钥加载到 openSSL 中?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我有这个 JWK(来自 https://datatracker.ietf.org/doc/html/draft-ietf-jose-json-web-signature-26#appendix-A.1):
{kty":RSA"，n":ofgWCuLjybRlzo0tZWJjNiuSfb4p4fAkd_wWJcyQoTbji9k0l8W26mPddxHmfHQp-Vaw-4qPCJrcS2mJPMEzP1Pt0Bm4d4QlL-yRT-SFd2lZS-pCgNMsD1W_YpRPEwOWvG6b32690r2jZ47soMZo9wGzjb_7OMg0LOL-bSf63kpaSHSXndS5z5rexMdbBYUsLA9e-KXBdQOS-UTo7WTBEMa2R2CapHg665xsmtdVMTBQY4uDZlxvb3qCo5ZwKh9kG4LT6_I5IhlJH7aGhyxXFvUK-DWNmoudF8NAco9_h9iaGNj8q2ethFkMLs91kzk2PAcDTW9gb54h4FRWyuXpoQ",e":AQAB"，d":Eq5xpGnNCivDflJsRQBXHx1hdR1k6Ulwe2JZD50LpXyWPEAeP88vLNO97IjlA7_GQ5sLKMgvfTeXZx9SE-7YwVol2NXOoAJe46sui395IW_GO-pWJ1O0BkTGoVen2bKVRUCgu-GjBVaYLU6f3l9kJfFNS3E0QbVdxzubSu3Mkqzjkn439X0M_V51gfpRLI9JYanrC4D4qAdGcopV_0ZHHzQlBjudU2QvXt4ehNYTCBr6XCLQUShb1juUO1ZdiYoFaFQT5Tw8bGUl_x_jTj3ccPDVZFD9pIuhLhBoneufuBiB4cS98l2SR_RQyGWSeWjnczT0QU91p1DhOVRuOopznQ";}我需要将它加载到一个 openSSL rsa 结构中，以便我可以将它输入到 EVP_SignFinal 调用中.d"是什么格式?质子交换膜?还是二进制?如何将其加载到 rsa 结构中?
解决方案
d"是什么格式?质子交换膜?还是二进制?
格式为Base64URL编码或Base 64 Encoding with URL and Filename Safe Alphabet"；RFC 4648(参见第 5 节，第 7 页的表 2).
<块引用>
如何将其加载到 rsa 结构中?
好的，所以 OpenSSL 很痛苦.要在 RSA 结构中加载它，您不需要从 Base64URL 转换 n、e 和 d 到 Base64.下面是我在 Crypto++ 中的做法(你可以在 OpenSSL 中这样做，但它会受到伤害):
string nz = ofgWCuLjybRlzo0tZWJjNiuSfb4p4fAkd_wWJcyQoTbji9k0l8W26mPddx"HmfHQp-Vaw-4qPCJrcS2mJPMEzP1Pt0Bm4d4QlL-yRT-SFd2lZS-pCgNMs"D1W_YpRPEwOWvG6b32690r2jZ47soMZo9wGzjb_7OMg0LOL-bSf63kpaSH"SXndS5z5rexMdbBYUsLA9e-KXBdQOS-UTo7WTBEMa2R2CapHg665xsmtdV"MTBQY4uDZlxvb3qCo5ZwKh9kG4LT6_I5IhlJH7aGhyxXFvUK-DWNmoudF8"NAco9_h9iaGNj8q2ethFkMLs91kzk2PAcDTW9gb54h4FRWyuXpoQ"；字符串 ez = "AQAB";字符串 dz = Eq5xpGnNCivDflJsRQBXHx1hdR1k6Ulwe2JZD50LpXyWPEAeP88vLNO97I"jlA7_GQ5sLKMgvfTeXZx9SE-7YwVol2NXOoAJe46sui395IW_GO-pWJ1O0"BkTGoVEn2bKVRUCgu-GjBVaYLU6f3l9kJfFNS3E0QbVdxzubSu3Mkqzjkn"439X0M_V51gfpRLI9JYanrC4D4qAdGcopV_0ZHHzQlBjudU2QvXt4ehNYT"CBr6XCLQUShb1juUO1ZdiYoFaFQT5Tw8bGUl_x_jTj3ccPDVZFD9pIuhLh"BOneufuBiB4cS98l2SR_RQyGWSeWjnczT0QU91p1DhOVRuOopznQ";字符串 nn，ee，dd；//首先将Base64URL编码转换为Base64std::replace(nz.begin(), nz.end(), '-', '+');std::replace(ez.begin(), ez.end(), '-', '+');std::replace(dz.begin(), dz.end(), '-', '+');std::replace(nz.begin(), nz.end(), '_', '/');std::replace(ez.begin(), ez.end(), '_', '/');std::replace(dz.begin(), dz.end(), '_', '/');//现在，Base64 解码StringSource ss1(nz, true, new Base64Decoder(new StringSink(nn)));StringSource ss2(ez, true, new Base64Decoder(new StringSink(ee)));StringSource ss3(dz, true, new Base64Decoder(new StringSink(dd)));编辑:Crypto++ 现在有 Base64URLEncoder 和 Base64URLDecoder 类，因此您不需要查找/替换操作.
上面的代码运行后，nn、ee和dd都是二进制字符串(即非ASCII字符).从那里，您可以将它们加载到 Integer 中并获取 Base 10 字符串:
整数 n((byte*)nn.data(), nn.size());整数 e((byte*)ee.data(), ee.size());整数 d((byte*)dd.data(), dd.size());cout <OpenSSL 需要 n、e、d、p 和 q 用于私钥操作.d mod p-1、d mod q-1 和 inv q mod p 是可选的.只有 n、e、d，您需要解决缺少的参数(至少 p 和 <代码>q).两个难点是 p 和 q.这是解决它们的 Crypto++ 代码(请随意转换为 OpenSSL):
整数 p, q;RSA_solve(n, e, d, p, q);cout <还有:
void RSA_solve(const Integer& n, const Integer& e, const Integer& d, Integer& p, Integer& q){AutoSeededRandomPool prng;整数 g = 1;无符号整数安全 = 0；步骤1:常量整数 k = e * d - 1;if(!k.IsEven())throw runtime_error("e * d - 1 is not even");第2步://g = 3, 5, 7, ...g += 2;while(!VerifyPrime(prng, g)) g += 2;整数 t = k;STEP_3:如果(安全++ > 128)throw runtime_error(不能考虑 n");如果(！t.IsEven())转到第 2 步；t/= 2；整数 x = a_exp_b_mod_c(g, t, n);步骤4:如果(！(x > 1))转到第 3 步；整数 y = GCD(x-1, n);if(!(y > 1))转到第 3 步；p = std::max(y, n/y);q = std::min(y, n/y);整数校验 = p * q;如果(n！=检查)抛出 runtime_error("n != p * q");}结果:
P:157377055902447438395586165028960291914931973278777532798470200156035267537359239071829408411909323208574959800537247728959718236884809685233284537349207654661530801859889389455120932077199406250387226339056140578989122526711937239401762061949364440402067108084155200696015505170135950332209194782224750221639.问:129921752567406358990993347540064445018230073402482260994179328573323861908379211274626956543471664997237185298964648133324343327052852264060322088122401124781249085873464824282666514908127141915943024862618996371026577302203267804867959037802770797169483022132210859867700312376409633383772189122488119155159.d mod p-1、d mod q-1 和 inv q mod p 留给读者作为练习(但它们很简单，尤其是在 Crypto++ 中).您修改后的 RSA_solve 可能如下所示:
void RSA_solve(const Integer& n, const Integer& e, const Integer& d,整数&p，整数&q,整数&dmodp1, 整数&dmodq1, 整数&invqmodp)现在，使用 Base 10(十进制)字符串切换到 OpenSSL:
const char nz[] =20446702916744654562596343388758805860065209639960173505037453331270270518732245"08977372301204320323609709562340204469011575537734525469644875960570778896584888"95017468362112062706438336639499925362469853626937363871851454247879222415857219"92924045675229348655595626434390043002821512765630397723028023792577935108185822"75369257422156693093780503115582009714681996492027000881132703628678639279359312"17624250488602118597634417704467037220158572506211078553986931332640811506974231"88751482418465308470313958250757758547155699749157985955379381294962058862159085"915015369381046959790476428631998204940879604226680285601";const char ez[] = "65537";常量字符 dz[] =23583109899396195101799862623499368829246520235662137651186064319555667005065389"11356936879137503597382515919515633242482643314423192704128296593672966061810149"31632061789402182278402640746140338406535182197235078430096761014345948432406842"76746396884059179774424728049430754391920261073195321175575450790865379829879825"22396626690057355718157403493216553255260857777965627529169195827622139772389760"13057175483467867984218114225248961766503010944557397801270779301059273764049922"00150833924259148778478404572782464027609558833769999511998277062853834711506435"61410605789710883438795588594095047409018233862167884701";常量字符 pz[] =15737705590244743839558616502896029191493197327877753279847020015603526753735923"90718294084119093232085749598005372477289597182368848096852332845373492076546615"30801859889389455120932077199406250387226339056140578989122526711937239401762061"949364440402067108084155200696015505170135950332209194782224750221639";常量字符 qz[] =12992175256740635899099334754006444501823007340248226099417932857332386190837921"12746269565434716649972371852989646481333243433270528522640603220881224011247812"49085873464824282666514908127141915943024862618996371026577302203267804867959037"802770797169483022132210859867700312376409633383772189122488119155159";使用 BN_ptr = std::unique_ptr;使用 RSA_ptr = std::unique_ptr;使用 EVP_PKEY_ptr = std::unique_ptr;使用 EVP_MD_CTX_ptr = std::unique_ptr;#define UNUSED(x) ((void)x)int main(int argc, char* argv[]){未使用(argc)；未使用(argv)；国际协调委员会；长错；RSA_ptr rsa(RSA_new(), ::RSA_free);BIGNUM *n = NULL，*e = NULL，*d = NULL，*p = NULL，*q = NULL；rc = BN_dec2bn(&n, nz);如果(rc == 0 || n == NULL){cerr <d = d;rc = BN_dec2bn(&p, pz);如果(rc == 0 || p == NULL){cerr <q = q;[练习留给读者]rc = RSA_check_key(rsa.get());错误 = ERR_get_error();如果(rc！= 1){cerr < 以下是您需要在 struct rsa 中提供的字段(来自
struct rsa_st{.../* 如果 'meth' 是 ENGINE 提供的，则函数引用 */发动机 *发动机；大数字 *n;大数字*e;大数字 *d;大数字 *p;大数字 *q;大数字 *dmp1;大数字 *dmq1;大数字 *iqmp;...};<块引用>
这样我就可以将它输入到 EVP_SignFinal 调用中...
EVP_SignFinal 需要一个 EVP_PKEY 并且你有一个 RSA.所以:
EVP_PKEY_ptr pkey(EVP_PKEY_new(), ::EVP_PKEY_free);rc = EVP_PKEY_set1_RSA(pkey.get(), rsa.get());错误 = ERR_get_error();如果(rc！= 1){cerr <set1 表示引用计数在 RSA* 上增加.没关系.如果它是 set0，您将不得不发布您的副本(即，使用 rsa.release() 而不是 rsa.get()) 以避免双重释放.
EVP_MD_CTX_ptr ctx(EVP_MD_CTX_create(), ::EVP_MD_CTX_destroy);EVP_MD_CTX_init(ctx.get());常量 EVP_MD* md = EVP_sha256();rc = EVP_SignInit(ctx.get(), md);错误 = ERR_get_error();如果(rc！= 1){cerr <签名(新的无符号字符[req])；无符号整数大小 = 请求；rc = EVP_SignFinal(ctx.get(), signature.get(), &size, pkey.get());错误 = ERR_get_error();如果(rc！= 1){cerr < 这是上面使用的 Crypto++ 代码的 Pastebin:http://pastebin.com/9Rm7bxZp.
这是上面使用的 OpenSSL 代码的 Pastebin:http://pastebin.com/aGVpj4FW.
这是 OpenSSL 程序的输出:
$ ./openssl-test.exe签名:78f2c9af23b9a2a42e3b57dec454fa43ea6627992f48d40a33da6a7c93f98b4I have this JWK (from https://datatracker.ietf.org/doc/html/draft-ietf-jose-json-web-signature-26#appendix-A.1):
{"kty":"RSA",
"n":"ofgWCuLjybRlzo0tZWJjNiuSfb4p4fAkd_wWJcyQoTbji9k0l8W26mPddx
HmfHQp-Vaw-4qPCJrcS2mJPMEzP1Pt0Bm4d4QlL-yRT-SFd2lZS-pCgNMs
D1W_YpRPEwOWvG6b32690r2jZ47soMZo9wGzjb_7OMg0LOL-bSf63kpaSH
SXndS5z5rexMdbBYUsLA9e-KXBdQOS-UTo7WTBEMa2R2CapHg665xsmtdV
MTBQY4uDZlxvb3qCo5ZwKh9kG4LT6_I5IhlJH7aGhyxXFvUK-DWNmoudF8
NAco9_h9iaGNj8q2ethFkMLs91kzk2PAcDTW9gb54h4FRWyuXpoQ",
"e":"AQAB",
"d":"Eq5xpGnNCivDflJsRQBXHx1hdR1k6Ulwe2JZD50LpXyWPEAeP88vLNO97I
jlA7_GQ5sLKMgvfTeXZx9SE-7YwVol2NXOoAJe46sui395IW_GO-pWJ1O0
BkTGoVEn2bKVRUCgu-GjBVaYLU6f3l9kJfFNS3E0QbVdxzubSu3Mkqzjkn
439X0M_V51gfpRLI9JYanrC4D4qAdGcopV_0ZHHzQlBjudU2QvXt4ehNYT
CBr6XCLQUShb1juUO1ZdiYoFaFQT5Tw8bGUl_x_jTj3ccPDVZFD9pIuhLh
BOneufuBiB4cS98l2SR_RQyGWSeWjnczT0QU91p1DhOVRuOopznQ"
}
I need to load it into an openSSL rsa struct, so that I can feed this into an EVP_SignFinal call. What format is "d"? PEM? or binary? how do I load it into a rsa struct?
解决方案
What format is "d"? PEM? or binary?
The format is Base64URL encoding or "Base 64 Encoding with URL and Filename Safe Alphabet" of RFC 4648 (see section 5, Table 2 on page 7).
how do I load it into a rsa struct?
OK, so OpenSSL is painful. To load it in a RSA struct, you need no convert n, e and d from Base64URL to Base64. Here's how I did it in Crypto++ (you can do it in OpenSSL, but its going to hurt):
string nz = "ofgWCuLjybRlzo0tZWJjNiuSfb4p4fAkd_wWJcyQoTbji9k0l8W26mPddx"
"HmfHQp-Vaw-4qPCJrcS2mJPMEzP1Pt0Bm4d4QlL-yRT-SFd2lZS-pCgNMs"
"D1W_YpRPEwOWvG6b32690r2jZ47soMZo9wGzjb_7OMg0LOL-bSf63kpaSH"
"SXndS5z5rexMdbBYUsLA9e-KXBdQOS-UTo7WTBEMa2R2CapHg665xsmtdV"
"MTBQY4uDZlxvb3qCo5ZwKh9kG4LT6_I5IhlJH7aGhyxXFvUK-DWNmoudF8"
"NAco9_h9iaGNj8q2ethFkMLs91kzk2PAcDTW9gb54h4FRWyuXpoQ";
string ez = "AQAB";
string dz = "Eq5xpGnNCivDflJsRQBXHx1hdR1k6Ulwe2JZD50LpXyWPEAeP88vLNO97I"
"jlA7_GQ5sLKMgvfTeXZx9SE-7YwVol2NXOoAJe46sui395IW_GO-pWJ1O0"
"BkTGoVEn2bKVRUCgu-GjBVaYLU6f3l9kJfFNS3E0QbVdxzubSu3Mkqzjkn"
"439X0M_V51gfpRLI9JYanrC4D4qAdGcopV_0ZHHzQlBjudU2QvXt4ehNYT"
"CBr6XCLQUShb1juUO1ZdiYoFaFQT5Tw8bGUl_x_jTj3ccPDVZFD9pIuhLh"
"BOneufuBiB4cS98l2SR_RQyGWSeWjnczT0QU91p1DhOVRuOopznQ";
string nn, ee, dd;
// First, convert Base64URL encoding to Base64
std::replace(nz.begin(), nz.end(), '-', '+');
std::replace(ez.begin(), ez.end(), '-', '+');
std::replace(dz.begin(), dz.end(), '-', '+');
std::replace(nz.begin(), nz.end(), '_', '/');
std::replace(ez.begin(), ez.end(), '_', '/');
std::replace(dz.begin(), dz.end(), '_', '/');
// Now, Base64 decode
StringSource ss1(nz, true, new Base64Decoder(new StringSink(nn)));
StringSource ss2(ez, true, new Base64Decoder(new StringSink(ee)));
StringSource ss3(dz, true, new Base64Decoder(new StringSink(dd)));
EDIT: Crypto++ now has the Base64URLEncoder and Base64URLDecoder classes, so you don't need the find/replace operations.
After the code above runs, nn, ee and dd are binary strings (i.e., non-ASCII characters). From there, you can load them into Integer's and get the Base 10 string with:
Integer n((byte*)nn.data(), nn.size());
Integer e((byte*)ee.data(), ee.size());
Integer d((byte*)dd.data(), dd.size());
cout << "N: " << endl << n << endl << endl;
cout << "E: " << endl << e << endl << endl;
cout << "D: " << endl << d << endl << endl;
$ ./cryptopp-test.exe
N:
20446702916744654562596343388758805860065209639960173505037453331270270518732245
08977372301204320323609709562340204469011575537734525469644875960570778896584888
95017468362112062706438336639499925362469853626937363871851454247879222415857219
92924045675229348655595626434390043002821512765630397723028023792577935108185822
75369257422156693093780503115582009714681996492027000881132703628678639279359312
17624250488602118597634417704467037220158572506211078553986931332640811506974231
88751482418465308470313958250757758547155699749157985955379381294962058862159085
915015369381046959790476428631998204940879604226680285601.
E:
65537.
D:
23583109899396195101799862623499368829246520235662137651186064319555667005065389
11356936879137503597382515919515633242482643314423192704128296593672966061810149
31632061789402182278402640746140338406535182197235078430096761014345948432406842
76746396884059179774424728049430754391920261073195321175575450790865379829879825
22396626690057355718157403493216553255260857777965627529169195827622139772389760
13057175483467867984218114225248961766503010944557397801270779301059273764049922
00150833924259148778478404572782464027609558833769999511998277062853834711506435
61410605789710883438795588594095047409018233862167884701.
OpenSSL needs n, e, d, p and q for private key operations. d mod p-1, d mod q-1 and inv q mod p are optional. With only n, e, d, you need to solve for the missing parameters (at minimum p and q). The two tough ones are p and q. Here's the Crypto++ code to solve for them (feel free to convert to OpenSSL):
Integer p, q;
RSA_solve(n, e, d, p, q);
cout << "P: " << endl << p << endl << endl;
cout << "Q: " << endl << q << endl << endl;
And:
void RSA_solve(const Integer& n, const Integer& e, const Integer& d, Integer& p, Integer& q)
{
AutoSeededRandomPool prng;
Integer g = 1;
unsigned int SAFETY = 0;
STEP_1:
const Integer k = e * d - 1;
if(!k.IsEven())
throw runtime_error("e * d - 1 is not even");
STEP_2:
// g = 3, 5, 7, ...
g += 2; while(!VerifyPrime(prng, g)) g += 2;
Integer t = k;
STEP_3:
if(SAFETY++ > 128)
throw runtime_error("could not factor n");
if(!t.IsEven())
goto STEP_2;
t /= 2;
Integer x = a_exp_b_mod_c(g, t, n);
STEP_4:
if(!(x > 1))
goto STEP_3;
Integer y = GCD(x-1, n);
if(!(y > 1))
goto STEP_3;
p = std::max(y, n/y);
q = std::min(y, n/y);
Integer check = p * q;
if(n != check)
throw runtime_error("n != p * q");
}
That results in:
P:
15737705590244743839558616502896029191493197327877753279847020015603526753735923
90718294084119093232085749598005372477289597182368848096852332845373492076546615
30801859889389455120932077199406250387226339056140578989122526711937239401762061
949364440402067108084155200696015505170135950332209194782224750221639.
Q:
12992175256740635899099334754006444501823007340248226099417932857332386190837921
12746269565434716649972371852989646481333243433270528522640603220881224011247812
49085873464824282666514908127141915943024862618996371026577302203267804867959037
802770797169483022132210859867700312376409633383772189122488119155159.
d mod p-1, d mod q-1 and inv q mod p are left as an exercise to the reader (but they are easy, especially in Crypto++). Your modified RSA_solve might look like:
void RSA_solve(const Integer& n, const Integer& e, const Integer& d,
Integer& p, Integer& q,
Integer& dmodp1, Integer& dmodq1, Integer& invqmodp)
Now, switch to OpenSSL with your Base 10 (decimal) strings:
const char nz[] =
"20446702916744654562596343388758805860065209639960173505037453331270270518732245"
"08977372301204320323609709562340204469011575537734525469644875960570778896584888"
"95017468362112062706438336639499925362469853626937363871851454247879222415857219"
"92924045675229348655595626434390043002821512765630397723028023792577935108185822"
"75369257422156693093780503115582009714681996492027000881132703628678639279359312"
"17624250488602118597634417704467037220158572506211078553986931332640811506974231"
"88751482418465308470313958250757758547155699749157985955379381294962058862159085"
"915015369381046959790476428631998204940879604226680285601";
const char ez[] = "65537";
const char dz[] =
"23583109899396195101799862623499368829246520235662137651186064319555667005065389"
"11356936879137503597382515919515633242482643314423192704128296593672966061810149"
"31632061789402182278402640746140338406535182197235078430096761014345948432406842"
"76746396884059179774424728049430754391920261073195321175575450790865379829879825"
"22396626690057355718157403493216553255260857777965627529169195827622139772389760"
"13057175483467867984218114225248961766503010944557397801270779301059273764049922"
"00150833924259148778478404572782464027609558833769999511998277062853834711506435"
"61410605789710883438795588594095047409018233862167884701";
const char pz[] =
"15737705590244743839558616502896029191493197327877753279847020015603526753735923"
"90718294084119093232085749598005372477289597182368848096852332845373492076546615"
"30801859889389455120932077199406250387226339056140578989122526711937239401762061"
"949364440402067108084155200696015505170135950332209194782224750221639";
const char qz[] =
"12992175256740635899099334754006444501823007340248226099417932857332386190837921"
"12746269565434716649972371852989646481333243433270528522640603220881224011247812"
"49085873464824282666514908127141915943024862618996371026577302203267804867959037"
"802770797169483022132210859867700312376409633383772189122488119155159";
using BN_ptr = std::unique_ptr<BIGNUM, decltype(&::BN_free)>;
using RSA_ptr = std::unique_ptr<RSA, decltype(&::RSA_free)>;
using EVP_PKEY_ptr = std::unique_ptr<EVP_PKEY, decltype(&::EVP_PKEY_free)>;
using EVP_MD_CTX_ptr = std::unique_ptr<EVP_MD_CTX, decltype(&::EVP_MD_CTX_destroy)>;
#define UNUSED(x) ((void)x)
int main(int argc, char* argv[])
{
UNUSED(argc); UNUSED(argv);
int rc;
long err;
RSA_ptr rsa(RSA_new(), ::RSA_free);
BIGNUM *n = NULL, *e = NULL, *d = NULL, *p = NULL, *q = NULL;
rc = BN_dec2bn(&n, nz);
if(rc == 0 || n == NULL) {
cerr << "BN_dec2bn failed for n" << endl;
exit(1);
}
rsa->n = n;
rc = BN_dec2bn(&e, ez);
if(rc == 0 || e == NULL) {
cerr << "BN_dec2bn failed for e" << endl;
exit(1);
}
rsa->e = e;
rc = BN_dec2bn(&d, dz);
if(rc == 0 || d == NULL) {
cerr << "BN_dec2bn failed for d" << endl;
exit(1);
}
rsa->d = d;
rc = BN_dec2bn(&p, pz);
if(rc == 0 || p == NULL) {
cerr << "BN_dec2bn failed for p" << endl;
exit(1);
}
rsa->p = p;
rc = BN_dec2bn(&q, qz);
if(rc == 0 || q == NULL) {
cerr << "BN_dec2bn failed for q" << endl;
exit(1);
}
rsa->q = q;
[Exercise left to the reader]
rc = RSA_check_key(rsa.get());
err = ERR_get_error();
if(rc != 1) {
cerr << "RSA_check_key failed, error 0x" << std::hex << err << endl;
exit(1);
}
[Continues at next question below]
...
}
Here are the fields you need to provide in struct rsa (from <openssl dir>/crypto/rsa/rsa.h):
struct rsa_st
{
...
/* functional reference if 'meth' is ENGINE-provided */
ENGINE *engine;
BIGNUM *n;
BIGNUM *e;
BIGNUM *d;
BIGNUM *p;
BIGNUM *q;
BIGNUM *dmp1;
BIGNUM *dmq1;
BIGNUM *iqmp;
...
};
so that I can feed this into an EVP_SignFinal call...
EVP_SignFinal needs an EVP_PKEY and you have an RSA. So:
EVP_PKEY_ptr pkey(EVP_PKEY_new(), ::EVP_PKEY_free);
rc = EVP_PKEY_set1_RSA(pkey.get(), rsa.get());
err = ERR_get_error();
if(rc != 1) {
cerr << "EVP_PKEY_set1_RSA failed, error 0x" << std::hex << err << endl;
exit(1);
}
The set1 means the reference count is bumped on the RSA*. That's OK. If it s was set0, you would have had to release your copy (that is, use rsa.release() rather than rsa.get()) to avoid a double free.
EVP_MD_CTX_ptr ctx(EVP_MD_CTX_create(), ::EVP_MD_CTX_destroy);
EVP_MD_CTX_init(ctx.get());
const EVP_MD* md = EVP_sha256();
rc = EVP_SignInit(ctx.get(), md);
err = ERR_get_error();
if(rc != 1) {
cerr << "EVP_SignInit_ex failed, error 0x" << std::hex << err << endl;
exit(1);
}
const char message[] = "Now is the time for all good men...";
rc = EVP_SignUpdate(ctx.get(), message, (unsigned int)sizeof(message));
err = ERR_get_error();
if(rc != 1) {
cerr << "EVP_SignUpdate failed, error 0x" << std::hex << err << endl;
exit(1);
}
const unsigned int req = std::max(EVP_MD_size(md), EVP_PKEY_size(pkey.get()));
unique_ptr<unsigned char[]> signature(new unsigned char[req]);
unsigned int size = req;
rc = EVP_SignFinal(ctx.get(), signature.get(), &size, pkey.get());
err = ERR_get_error();
if(rc != 1) {
cerr << "EVP_SignFinal failed, error 0x" << std::hex << err << endl;
exit(1);
}
size = std::min(size, (unsigned int)EVP_MD_size(md));
cout << "Signature: ";
for(unsigned i = 0; i < size; i++)
cout << std::hex << (signature[i] & 0xFF);
cout << endl;
Here's the Pastebin of the Crypto++ code used above: http://pastebin.com/9Rm7bxZp.
Here's the Pastebin of the OpenSSL code used above: http://pastebin.com/aGVpj4FW.
Here's the output of the OpenSSL program:
$ ./openssl-test.exe
Signature: 78f2c9af23b9a2a42e3b57dec454fa43ea6627992f48d40a33da6a7c93f98b4
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