我正在尝试在C中实现AWS V4 signature，但未能计算要签名的字符串中的哈希。测试示例here具有：
key = 'wJalrXUtnFEMI/K7MDENG+bPxRfiCYEXAMPLEKEY'
dateStamp = '20120215'
regionName = 'us-east-1'
serviceName = 'iam'
生产
kSecret = '41575334774a616c725855746e46454d492f4b374d44454e472b62507852666943594558414d504c454b4559'
kDate = '969fbb94feb542b71ede6f87fe4d5fa29c789342b0f407474670f0c2489e0a0d'
kRegion = '69daa0209cd9c5ff5c8ced464a696fd4252e981430b10e3d3fd8e2f197d7a70c'
kService = 'f72cfd46f26bc4643f06a11eabb6c0ba18780c19a8da0c31ace671265e3c87fa'
kSigning = 'f4780e2d9f65fa895f9c67b32ce1baf0b0d8a43505a000a1a9e090d414db404d'
但是，由于某种原因，我的实现删除了final元素的最后几个字符，我得到了：
kSecret : 41575334774a616c725855746e46454d492f4b374d44454e472b62507852666943594558414d504c454b4559
kDate : 969fbb94feb542b71ede6f87fe4d5fa29c789342b0f407474670f0c2489e0a0d
kRegion : 69daa0209cd9c5ff5c8ced464a696fd4252e981430b10e3d3fd8e2f197d7a70c
kService : f72cfd46f26bc4643f06a11eabb6c0ba18780c19a8da0c31ace671265e3c87fa
kSigning : f4780e2d9f65fa895f9c67b32ce1baf0b0d8a43505a0
我只是使用来自https://stackoverflow.com/a/29862424/993874的HMAC_SHA256计算示例，我看不到我所缺少的内容。对于其他计算（即对我自己的AWS数据执行请求的计算等），我还看到某些hmac输出的长度不是32个字符。
任何人都可以帮助解决这个问题（并确认是否需要将所有char数组转换为针对AWS请求的十六进制字符串）？
示例代码：
#include <stdio.h>
#include <string.h>
#include <openssl/hmac.h>
void printHex(const char* preface, const char* toPrint) {
printf("%s\t: ", preface);
for(size_t i = 0; i < strlen(toPrint); i++) {
printf("%02x", toPrint[i] & 0xff);
}
printf("\n");
}
void hmac_sha256(const unsigned char* key,
const unsigned char* text,
unsigned char* result) {
unsigned int resultLen;
HMAC(EVP_sha256(), key, strlen(key), text, strlen(text), result, &resultLen);
}
void amazonV4Sign(const unsigned char* accessSecret,
const unsigned char* amzDate,
const unsigned char* region,
const unsigned char* service) {
unsigned char kDate[BUFSIZ];
unsigned char kRegion[BUFSIZ];
unsigned char kService[BUFSIZ];
unsigned char kSigning[BUFSIZ];
unsigned char kSecret[BUFSIZ];
sprintf(kSecret, "AWS4%s", accessSecret);
unsigned char request[BUFSIZ];
sprintf(request, "aws4_request");
hmac_sha256(kSecret, amzDate, kDate);
hmac_sha256(kDate, region, kRegion);
hmac_sha256(kRegion, service, kService);
hmac_sha256(kService, request, kSigning);
printHex("kSecret", kSecret);
printHex("kDate", kDate);
printHex("kRegion", kRegion);
printHex("kService", kService);
printHex("kSigning", kSigning);
}
int main(int argc, char* argv[]) {
unsigned char* key = "wJalrXUtnFEMI/K7MDENG+bPxRfiCYEXAMPLEKEY";
unsigned char* dateStamp = "20120215";
unsigned char* regionName = "us-east-1";
unsigned char* serviceName = "iam";
amazonV4Sign(key, dateStamp, regionName, serviceName);
return 0;
}
答案 0 :(得分：1)
您不能使用def applyConditionals[T](toApply: (() => Boolean, T => T)*) = toApply
.foldLeft(a) {
case (a, (cond, oper)) if cond() => oper(a)
case (a, _) => a
}
val result = applyConditionals[Int] (
(() => condition, _ * 2),
(() => condition2, _ * 6),
(() => condition3, _ * 8)
)
来打印strlen输出，因为字节数组可能包含HMAC个字节。
因此，您可以如下使用0返回的md_len来打印字节数组。
HMAC返回HMAC的长度，语法为。
md因此使用unsigned char *HMAC(const EVP_MD *evp_md, const void *key,
int key_len, const unsigned char *d, int n,
unsigned char *md, unsigned int *md_len);
打印数组。
md_len输出：
void printHex(const char* preface, const char* toPrint, int len) {
printf("%s\t: ", preface);
for(size_t i = 0; i < len; i++) {
printf("%02x", toPrint[i] & 0xff);
}
printf("\n");
}
int hmac_sha256(const unsigned char* key,
const unsigned char* text,
unsigned char* result) {
unsigned int resultLen;
HMAC(EVP_sha256(), key, strlen(key), text, strlen(text), result, &resultLen);
printf("%d\n", resultLen);
return resultLen;
}
void amazonV4Sign(const unsigned char* accessSecret,
const unsigned char* amzDate,
const unsigned char* region,
const unsigned char* service) {
unsigned char kDate[EVP_MAX_MD_SIZE] = {0};
unsigned char kRegion[EVP_MAX_MD_SIZE] = {0};
unsigned char kService[EVP_MAX_MD_SIZE] = {0};
unsigned char kSigning[EVP_MAX_MD_SIZE] = {0};
unsigned char kSecret[BUFSIZ];
sprintf(kSecret, "AWS4%s", accessSecret);
printf("%s\n", kSecret);
unsigned char request[BUFSIZ];
sprintf(request, "aws4_request");
int len = hmac_sha256(kSecret, amzDate, kDate);
printHex("kSecret", kSecret,strlen(kSecret));
printHex("kDate", kDate,len);
len = hmac_sha256(kDate, region, kRegion);
printHex("kRegion", kRegion,len);
len = hmac_sha256(kRegion, service, kService);
printHex("kService", kService,len);
len = hmac_sha256(kService, request, kSigning);
printHex("kSigning", kSigning,len);
}
答案 1 :(得分：0)
您在PrintHex中的循环最多为const app = dialogflow({debug: true, clientId:'*.apps.googleusercontent.com'});
var firebase = require('firebase');
const {dialogflow} = require('actions-on-google');
const functions = require('firebase-functions');
app.intent('Default Welcome Intent',(conv) =>{
conv.ask(new SignIn('To get your account details'));
});
app.intent('Get Signin', (conv, signin) => {
if (signin.status === 'OK') {
const payload = conv.user.profile.payload
conv.ask(`I got your account details, ${payload.name}. What do you want to do next?`)
} else {
conv.ask(`I won't be able to save your data, but what do you want to do next?`)
}
});
。它是一个字节数组，而不是字符串，因此您不能使用strlen（）。循环最多需要32个。