任何人都可以在c ++中免费获得base64解码代码片段吗？
答案 0 :(得分：88)
以下是我对RenéNyffenegger最初编写的implementation的修改。为什么我要修改它？好吧，因为我似乎不适合使用存储在std::string对象中的二进制数据;）
<强> base64.h ：
#ifndef _BASE64_H_
#define _BASE64_H_
#include <vector>
#include <string>
typedef unsigned char BYTE;
std::string base64_encode(BYTE const* buf, unsigned int bufLen);
std::vector<BYTE> base64_decode(std::string const&);
#endif
<强> base64.cpp ：
#include "base64.h"
#include <iostream>
static const std::string base64_chars =
"ABCDEFGHIJKLMNOPQRSTUVWXYZ"
"abcdefghijklmnopqrstuvwxyz"
"0123456789+/";
static inline bool is_base64(BYTE c) {
return (isalnum(c) || (c == '+') || (c == '/'));
}
std::string base64_encode(BYTE const* buf, unsigned int bufLen) {
std::string ret;
int i = 0;
int j = 0;
BYTE char_array_3[3];
BYTE char_array_4[4];
while (bufLen--) {
char_array_3[i++] = *(buf++);
if (i == 3) {
char_array_4[0] = (char_array_3[0] & 0xfc) >> 2;
char_array_4[1] = ((char_array_3[0] & 0x03) << 4) + ((char_array_3[1] & 0xf0) >> 4);
char_array_4[2] = ((char_array_3[1] & 0x0f) << 2) + ((char_array_3[2] & 0xc0) >> 6);
char_array_4[3] = char_array_3[2] & 0x3f;
for(i = 0; (i <4) ; i++)
ret += base64_chars[char_array_4[i]];
i = 0;
}
}
if (i)
{
for(j = i; j < 3; j++)
char_array_3[j] = '\0';
char_array_4[0] = (char_array_3[0] & 0xfc) >> 2;
char_array_4[1] = ((char_array_3[0] & 0x03) << 4) + ((char_array_3[1] & 0xf0) >> 4);
char_array_4[2] = ((char_array_3[1] & 0x0f) << 2) + ((char_array_3[2] & 0xc0) >> 6);
char_array_4[3] = char_array_3[2] & 0x3f;
for (j = 0; (j < i + 1); j++)
ret += base64_chars[char_array_4[j]];
while((i++ < 3))
ret += '=';
}
return ret;
}
std::vector<BYTE> base64_decode(std::string const& encoded_string) {
int in_len = encoded_string.size();
int i = 0;
int j = 0;
int in_ = 0;
BYTE char_array_4[4], char_array_3[3];
std::vector<BYTE> ret;
while (in_len-- && ( encoded_string[in_] != '=') && is_base64(encoded_string[in_])) {
char_array_4[i++] = encoded_string[in_]; in_++;
if (i ==4) {
for (i = 0; i <4; i++)
char_array_4[i] = base64_chars.find(char_array_4[i]);
char_array_3[0] = (char_array_4[0] << 2) + ((char_array_4[1] & 0x30) >> 4);
char_array_3[1] = ((char_array_4[1] & 0xf) << 4) + ((char_array_4[2] & 0x3c) >> 2);
char_array_3[2] = ((char_array_4[2] & 0x3) << 6) + char_array_4[3];
for (i = 0; (i < 3); i++)
ret.push_back(char_array_3[i]);
i = 0;
}
}
if (i) {
for (j = i; j <4; j++)
char_array_4[j] = 0;
for (j = 0; j <4; j++)
char_array_4[j] = base64_chars.find(char_array_4[j]);
char_array_3[0] = (char_array_4[0] << 2) + ((char_array_4[1] & 0x30) >> 4);
char_array_3[1] = ((char_array_4[1] & 0xf) << 4) + ((char_array_4[2] & 0x3c) >> 2);
char_array_3[2] = ((char_array_4[2] & 0x3) << 6) + char_array_4[3];
for (j = 0; (j < i - 1); j++) ret.push_back(char_array_3[j]);
}
return ret;
}
以下是用法：
std::vector<BYTE> myData;
...
std::string encodedData = base64_encode(&myData[0], myData.size());
std::vector<BYTE> decodedData = base64_decode(encodedData);
我希望有人会发现这个答案有用^^
答案 1 :(得分：73)
以下是该页面的实施：
/*
base64.cpp and base64.h
Copyright (C) 2004-2008 René Nyffenegger
This source code is provided 'as-is', without any express or implied
warranty. In no event will the author be held liable for any damages
arising from the use of this software.
Permission is granted to anyone to use this software for any purpose,
including commercial applications, and to alter it and redistribute it
freely, subject to the following restrictions:
1. The origin of this source code must not be misrepresented; you must not
claim that you wrote the original source code. If you use this source code
in a product, an acknowledgment in the product documentation would be
appreciated but is not required.
2. Altered source versions must be plainly marked as such, and must not be
misrepresented as being the original source code.
3. This notice may not be removed or altered from any source distribution.
René Nyffenegger rene.nyffenegger@adp-gmbh.ch
*/
static const std::string base64_chars =
"ABCDEFGHIJKLMNOPQRSTUVWXYZ"
"abcdefghijklmnopqrstuvwxyz"
"0123456789+/";
static inline bool is_base64(unsigned char c) {
return (isalnum(c) || (c == '+') || (c == '/'));
}
std::string base64_encode(unsigned char const* bytes_to_encode, unsigned int in_len) {
std::string ret;
int i = 0;
int j = 0;
unsigned char char_array_3[3];
unsigned char char_array_4[4];
while (in_len--) {
char_array_3[i++] = *(bytes_to_encode++);
if (i == 3) {
char_array_4[0] = (char_array_3[0] & 0xfc) >> 2;
char_array_4[1] = ((char_array_3[0] & 0x03) << 4) + ((char_array_3[1] & 0xf0) >> 4);
char_array_4[2] = ((char_array_3[1] & 0x0f) << 2) + ((char_array_3[2] & 0xc0) >> 6);
char_array_4[3] = char_array_3[2] & 0x3f;
for(i = 0; (i <4) ; i++)
ret += base64_chars[char_array_4[i]];
i = 0;
}
}
if (i)
{
for(j = i; j < 3; j++)
char_array_3[j] = '\0';
char_array_4[0] = (char_array_3[0] & 0xfc) >> 2;
char_array_4[1] = ((char_array_3[0] & 0x03) << 4) + ((char_array_3[1] & 0xf0) >> 4);
char_array_4[2] = ((char_array_3[1] & 0x0f) << 2) + ((char_array_3[2] & 0xc0) >> 6);
char_array_4[3] = char_array_3[2] & 0x3f;
for (j = 0; (j < i + 1); j++)
ret += base64_chars[char_array_4[j]];
while((i++ < 3))
ret += '=';
}
return ret;
}
std::string base64_decode(std::string const& encoded_string) {
int in_len = encoded_string.size();
int i = 0;
int j = 0;
int in_ = 0;
unsigned char char_array_4[4], char_array_3[3];
std::string ret;
while (in_len-- && ( encoded_string[in_] != '=') && is_base64(encoded_string[in_])) {
char_array_4[i++] = encoded_string[in_]; in_++;
if (i ==4) {
for (i = 0; i <4; i++)
char_array_4[i] = base64_chars.find(char_array_4[i]);
char_array_3[0] = (char_array_4[0] << 2) + ((char_array_4[1] & 0x30) >> 4);
char_array_3[1] = ((char_array_4[1] & 0xf) << 4) + ((char_array_4[2] & 0x3c) >> 2);
char_array_3[2] = ((char_array_4[2] & 0x3) << 6) + char_array_4[3];
for (i = 0; (i < 3); i++)
ret += char_array_3[i];
i = 0;
}
}
if (i) {
for (j = i; j <4; j++)
char_array_4[j] = 0;
for (j = 0; j <4; j++)
char_array_4[j] = base64_chars.find(char_array_4[j]);
char_array_3[0] = (char_array_4[0] << 2) + ((char_array_4[1] & 0x30) >> 4);
char_array_3[1] = ((char_array_4[1] & 0xf) << 4) + ((char_array_4[2] & 0x3c) >> 2);
char_array_3[2] = ((char_array_4[2] & 0x3) << 6) + char_array_4[3];
for (j = 0; (j < i - 1); j++) ret += char_array_3[j];
}
return ret;
}
答案 2 :(得分：28)
这里有几个片段。然而，这个是紧凑，高效，和c ++ 11友好：
static std::string base64_encode(const std::string &in) {
std::string out;
int val=0, valb=-6;
for (uchar c : in) {
val = (val<<8) + c;
valb += 8;
while (valb>=0) {
out.push_back("ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/"[(val>>valb)&0x3F]);
valb-=6;
}
}
if (valb>-6) out.push_back("ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/"[((val<<8)>>(valb+8))&0x3F]);
while (out.size()%4) out.push_back('=');
return out;
}
static std::string base64_decode(const std::string &in) {
std::string out;
std::vector<int> T(256,-1);
for (int i=0; i<64; i++) T["ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/"[i]] = i;
int val=0, valb=-8;
for (uchar c : in) {
if (T[c] == -1) break;
val = (val<<6) + T[c];
valb += 6;
if (valb>=0) {
out.push_back(char((val>>valb)&0xFF));
valb-=8;
}
}
return out;
}
答案 3 :(得分：13)
使用base-n迷你lib，您可以执行以下操作：
some_data_t in[] { ... };
constexpr int len = sizeof(in)/sizeof(in[0]);
std::string encoded;
bn::encode_b64(in, in + len, std::back_inserter(encoded));
some_data_t out[len];
bn::decode_b64(encoded.begin(), encoded.end(), out);
API是通用的，基于迭代器的。
披露：我是作者。
答案 4 :(得分：10)
我认为这个效果更好：
#include <string>
static const char* B64chars = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
static const int B64index[256] =
{
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 62, 63, 62, 62, 63,
52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 0, 0, 0, 0, 0, 0,
0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14,
15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 0, 0, 0, 0, 63,
0, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40,
41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51
};
const std::string b64encode(const void* data, const size_t &len)
{
std::string result((len + 2) / 3 * 4, '=');
char *p = (char*) data, *str = &result[0];
size_t j = 0, pad = len % 3;
const size_t last = len - pad;
for (size_t i = 0; i < last; i += 3)
{
int n = int(p[i]) << 16 | int(p[i + 1]) << 8 | p[i + 2];
str[j++] = B64chars[n >> 18];
str[j++] = B64chars[n >> 12 & 0x3F];
str[j++] = B64chars[n >> 6 & 0x3F];
str[j++] = B64chars[n & 0x3F];
}
if (pad) /// set padding
{
int n = --pad ? int(p[last]) << 8 | p[last + 1] : p[last];
str[j++] = B64chars[pad ? n >> 10 & 0x3F : n >> 2];
str[j++] = B64chars[pad ? n >> 4 & 0x03F : n << 4 & 0x3F];
str[j++] = pad ? B64chars[n << 2 & 0x3F] : '=';
}
return result;
}
const std::string b64decode(const void* data, const size_t &len)
{
if (len == 0) return "";
unsigned char *p = (unsigned char*) data;
size_t j = 0,
pad1 = len % 4 || p[len - 1] == '=',
pad2 = pad1 && (len % 4 > 2 || p[len - 2] != '=');
const size_t last = (len - pad1) / 4 << 2;
std::string result(last / 4 * 3 + pad1 + pad2, '\0');
unsigned char *str = (unsigned char*) &result[0];
for (size_t i = 0; i < last; i += 4)
{
int n = B64index[p[i]] << 18 | B64index[p[i + 1]] << 12 | B64index[p[i + 2]] << 6 | B64index[p[i + 3]];
str[j++] = n >> 16;
str[j++] = n >> 8 & 0xFF;
str[j++] = n & 0xFF;
}
if (pad1)
{
int n = B64index[p[last]] << 18 | B64index[p[last + 1]] << 12;
str[j++] = n >> 16;
if (pad2)
{
n |= B64index[p[last + 2]] << 6;
str[j++] = n >> 8 & 0xFF;
}
}
return result;
}
std::string b64encode(const std::string& str)
{
return b64encode(str.c_str(), str.size());
}
std::string b64decode(const std::string& str64)
{
return b64decode(str64.c_str(), str64.size());
}
感谢@Jens Alfke指出性能问题，我对此旧帖子进行了一些修改。这个比以前更快。它的另一个优点是可以顺利处理损坏的数据。
上一版：虽然在这些类型的问题中，似乎速度是一种矫枉过正，但仅仅为了它的乐趣，我做了一些其他的修改，使这一个成为最快的算法据我所知。特别感谢@GaspardP提出的宝贵建议和不错的基准。
答案 5 :(得分：9)
根据this GaspardP的优异比较，我不会选择这个解决方案。这不是最糟糕的，但它也不是最好的。它唯一能做的就是它可能更容易理解。
我发现另外两个答案很难理解。它们还会在我的编译器中产生一些警告，并且在解码部分中使用find函数会导致效率非常差。所以我决定自己动手。
部首：
--all体：
#ifndef _BASE64_H_
#define _BASE64_H_
#include <vector>
#include <string>
typedef unsigned char BYTE;
class Base64
{
public:
static std::string encode(const std::vector<BYTE>& buf);
static std::string encode(const BYTE* buf, unsigned int bufLen);
static std::vector<BYTE> decode(std::string encoded_string);
};
#endif
用法：
static const BYTE from_base64[] = { 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255,
255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255,
255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 62, 255, 62, 255, 63,
52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 255, 255, 255, 255, 255, 255,
255, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14,
15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 255, 255, 255, 255, 63,
255, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40,
41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 255, 255, 255, 255, 255};
static const char to_base64[] =
"ABCDEFGHIJKLMNOPQRSTUVWXYZ"
"abcdefghijklmnopqrstuvwxyz"
"0123456789+/";
std::string Base64::encode(const std::vector<BYTE>& buf)
{
if (buf.empty())
return ""; // Avoid dereferencing buf if it's empty
return encode(&buf[0], (unsigned int)buf.size());
}
std::string Base64::encode(const BYTE* buf, unsigned int bufLen)
{
// Calculate how many bytes that needs to be added to get a multiple of 3
size_t missing = 0;
size_t ret_size = bufLen;
while ((ret_size % 3) != 0)
{
++ret_size;
++missing;
}
// Expand the return string size to a multiple of 4
ret_size = 4*ret_size/3;
std::string ret;
ret.reserve(ret_size);
for (unsigned int i=0; i<ret_size/4; ++i)
{
// Read a group of three bytes (avoid buffer overrun by replacing with 0)
size_t index = i*3;
BYTE b3[3];
b3[0] = (index+0 < bufLen) ? buf[index+0] : 0;
b3[1] = (index+1 < bufLen) ? buf[index+1] : 0;
b3[2] = (index+2 < bufLen) ? buf[index+2] : 0;
// Transform into four base 64 characters
BYTE b4[4];
b4[0] = ((b3[0] & 0xfc) >> 2);
b4[1] = ((b3[0] & 0x03) << 4) + ((b3[1] & 0xf0) >> 4);
b4[2] = ((b3[1] & 0x0f) << 2) + ((b3[2] & 0xc0) >> 6);
b4[3] = ((b3[2] & 0x3f) << 0);
// Add the base 64 characters to the return value
ret.push_back(to_base64[b4[0]]);
ret.push_back(to_base64[b4[1]]);
ret.push_back(to_base64[b4[2]]);
ret.push_back(to_base64[b4[3]]);
}
// Replace data that is invalid (always as many as there are missing bytes)
for (size_t i=0; i<missing; ++i)
ret[ret_size - i - 1] = '=';
return ret;
}
std::vector<BYTE> Base64::decode(std::string encoded_string)
{
// Make sure string length is a multiple of 4
while ((encoded_string.size() % 4) != 0)
encoded_string.push_back('=');
size_t encoded_size = encoded_string.size();
std::vector<BYTE> ret;
ret.reserve(3*encoded_size/4);
for (size_t i=0; i<encoded_size; i += 4)
{
// Get values for each group of four base 64 characters
BYTE b4[4];
b4[0] = (encoded_string[i+0] <= 'z') ? from_base64[encoded_string[i+0]] : 0xff;
b4[1] = (encoded_string[i+1] <= 'z') ? from_base64[encoded_string[i+1]] : 0xff;
b4[2] = (encoded_string[i+2] <= 'z') ? from_base64[encoded_string[i+2]] : 0xff;
b4[3] = (encoded_string[i+3] <= 'z') ? from_base64[encoded_string[i+3]] : 0xff;
// Transform into a group of three bytes
BYTE b3[3];
b3[0] = ((b4[0] & 0x3f) << 2) + ((b4[1] & 0x30) >> 4);
b3[1] = ((b4[1] & 0x0f) << 4) + ((b4[2] & 0x3c) >> 2);
b3[2] = ((b4[2] & 0x03) << 6) + ((b4[3] & 0x3f) >> 0);
// Add the byte to the return value if it isn't part of an '=' character (indicated by 0xff)
if (b4[1] != 0xff) ret.push_back(b3[0]);
if (b4[2] != 0xff) ret.push_back(b3[1]);
if (b4[3] != 0xff) ret.push_back(b3[2]);
}
return ret;
}
这里的一个好处是解码函数还可以解码base 64编码的url变体。
答案 6 :(得分：4)
我对@DaedalusAlpha的变化回答。 它避免了以几次测试为代价来复制参数。
使用uint8_t而不是BYTE。
添加一些方便的函数来处理字符串，虽然输入数据通常是二进制的，内部可能没有字节，因此通常不应该被操作为字符串（通常意味着以空值终止的数据）。
还添加了一些强制转换来修复编译器警告（至少在GCC上，我还没有通过MSVC运行它）。
base64.hpp的一部分：
$res=$test_buck->upsert("cache_test1", '{"attr": "ran"}');
base64.cpp：
void base64_encode(string & out, const vector<uint8_t>& buf);
void base64_encode(string & out, const uint8_t* buf, size_t bufLen);
void base64_encode(string & out, string const& buf);
void base64_decode(vector<uint8_t> & out, string const& encoded_string);
// use this if you know the output should be a valid string
void base64_decode(string & out, string const& encoded_string);
答案 7 :(得分：2)
使用更紧凑的查找表和使用c ++ 17功能的一点变化：
std::string base64_decode(const std::string_view in) {
// table from '+' to 'z'
const uint8_t lookup[] = {
62, 255, 62, 255, 63, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 255,
255, 0, 255, 255, 255, 255, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25,
255, 255, 255, 255, 63, 255, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35,
36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51};
static_assert(sizeof(lookup) == 'z' - '+' + 1);
std::string out;
int val = 0, valb = -8;
for (uint8_t c : in) {
if (c < '+' || c > 'z')
break;
c -= '+';
if (lookup[c] >= 64)
break;
val = (val << 6) + lookup[c];
valb += 6;
if (valb >= 0) {
out.push_back(char((val >> valb) & 0xFF));
valb -= 8;
}
}
return out;
}
如果您没有std :: string_view，请尝试使用std :: experimental :: string_view。
答案 8 :(得分：1)
我的版本是C ++ Builder的base64的简单快速编码器（解码器）。
//---------------------------------------------------------------------------
UnicodeString __fastcall TExample::Base64Encode(void *data,int length)
{
if (length<=0) return L"";
static const char set[]="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
unsigned char *in=(unsigned char*)data;
char *pos,*out=pos=new char[((length-1)/3+1)<<2];
while ((length-=3)>=0)
{
pos[0]=set[in[0]>>2];
pos[1]=set[((in[0]&0x03)<<4)|(in[1]>>4)];
pos[2]=set[((in[1]&0x0F)<<2)|(in[2]>>6)];
pos[3]=set[in[2]&0x3F];
pos+=4;
in+=3;
};
if ((length&2)!=0)
{
pos[0]=set[in[0]>>2];
if ((length&1)!=0)
{
pos[1]=set[((in[0]&0x03)<<4)|(in[1]>>4)];
pos[2]=set[(in[1]&0x0F)<<2];
}
else
{
pos[1]=set[(in[0]&0x03)<<4];
pos[2]='=';
};
pos[3]='=';
pos+=4;
};
UnicodeString code=UnicodeString(out,pos-out);
delete[] out;
return code;
};
//---------------------------------------------------------------------------
int __fastcall TExample::Base64Decode(const UnicodeString code,unsigned char **data)
{
int length;
if (((length=code.Length())==0)||((length&3)!=0)) return 0;
wchar_t *str=code.c_str();
unsigned char *pos,*out=pos=new unsigned char[(length>>2)*3];
while (*str!=0)
{
length=-1;
int shift=18,bits=0;
do
{
wchar_t s=str[++length];
if ((s>=L'A')&&(s<=L'Z')) bits|=(s-L'A')<<shift;
else if ((s>=L'a')&&(s<=L'z')) bits|=(s-(L'a'-26))<<shift;
else if (((s>=L'0')&&(s<=L'9'))) bits|=(s-(L'0'-52))<<shift;
else if (s==L'+') bits|=62<<shift;
else if (s==L'/') bits|=63<<shift;
else if (s==L'=')
{
length--;
break;
}
else
{
delete[] out;
return 0;
};
}
while ((shift-=6)>=0);
pos[0]=bits>>16;
pos[1]=bits>>8;
pos[2]=bits;
pos+=length;
str+=4;
};
*data=out;
return pos-out;
};
//---------------------------------------------------------------------------
答案 9 :(得分：0)
轮到我了。我用这个：
class BinaryVector {
public:
std::vector<char> bytes;
uint64_t bit_count = 0;
public:
/* add a bit to the end */
void push_back(bool bit);
/* return false if character is unrecognized */
bool pushBase64Char(char b64_c);
};
void bin::Vector::push_back(bool bit)
{
if (!bit_count || bit_count % 8 == 0) {
bytes.push_back(bit << 7);
}
else {
uint8_t next_bit = 8 - (bit_count % 8) - 1;
bytes[bit_count / 8] |= bit << next_bit;
}
bit_count++;
}
/* converts one Base64 character to 6 bits */
bool bin::Vector::pushBase64Char(char c)
{
uint8_t d;
// A to Z
if (c > 0x40 && c < 0x5b) {
d = c - 65; // Base64 A is 0
}
// a to z
else if (c > 0x60 && c < 0x7b) {
d = c - 97 + 26; // Base64 a is 26
}
// 0 to 9
else if (c > 0x2F && c < 0x3a) {
d = c - 48 + 52; // Base64 0 is 52
}
else if (c == '+') {
d = 0b111110;
}
else if (c == '/') {
d = 0b111111;
}
else if (c == '=') {
d = 0;
}
else {
return false;
}
push_back(d & 0b100000);
push_back(d & 0b010000);
push_back(d & 0b001000);
push_back(d & 0b000100);
push_back(d & 0b000010);
push_back(d & 0b000001);
return true;
}
bool loadBase64(std::vector<char>& b64_bin, BinaryVector& vec)
{
for (char& c : b64_bin) {
if (!vec.pushBase64Char(c)) {
return false;
}
}
return true;
}
使用vec.bytes访问转换后的数据。
答案 10 :(得分：0)
我首先制作了自己的版本，然后找到了这个主题。
为什么我的版本看起来比这里介绍的其他版本简单？难道我做错了什么？我没有测试速度。
inline char const* b64units = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
inline char* b64encode(void const* a, int64_t b) {
ASSERT(a != nullptr);
if (b > 0) {
uint8_t const* aa = static_cast<uint8_t const*>(a);
uint8_t v = 0;
int64_t bp = 0;
int64_t sb = 0;
int8_t off = 0;
int64_t nt = ((b + 2) / 3) * 4;
int64_t nd = (b * 8) / 6;
int64_t tl = ((b * 8) % 6) ? 1 : 0;
int64_t nf = nt - nd - tl;
int64_t ri = 0;
char* r = new char[nt + 1]();
for (int64_t i = 0; i < nd; i++) {
v = (aa[sb] << off) | (aa[sb + 1] >> (8 - off));
v >>= 2;
r[ri] = b64units[v];
ri += 1;
bp += 6;
sb = (bp / 8);
off = (bp % 8);
}
if (tl > 0) {
v = (aa[sb] << off);
v >>= 2;
r[ri] = b64units[v];
ri += 1;
}
for (int64_t i = 0; i < nf; i++) {
r[ri] = '=';
ri += 1;
}
return r;
} else return nullptr;
}
P.S。 我的方法效果很好，我使用Node.js进行了测试：
let data = 'stackabuse.com';
let buff = new Buffer(data);
let base64data = buff.toString('base64');
答案 11 :(得分：0)
它是一个单一的 hpp 文件，与接受的答案不同，它使用 vector
#pragma once
#include <string>
#include <vector>
#include <stdexcept>
#include <cstdint>
namespace base64
{
inline static const char kEncodeLookup[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
inline static const char kPadCharacter = '=';
using byte = std::uint8_t;
inline std::string encode(const std::vector<byte>& input)
{
std::string encoded;
encoded.reserve(((input.size() / 3) + (input.size() % 3 > 0)) * 4);
std::uint32_t temp{};
auto it = input.begin();
for(std::size_t i = 0; i < input.size() / 3; ++i)
{
temp = (*it++) << 16;
temp += (*it++) << 8;
temp += (*it++);
encoded.append(1, kEncodeLookup[(temp & 0x00FC0000) >> 18]);
encoded.append(1, kEncodeLookup[(temp & 0x0003F000) >> 12]);
encoded.append(1, kEncodeLookup[(temp & 0x00000FC0) >> 6 ]);
encoded.append(1, kEncodeLookup[(temp & 0x0000003F) ]);
}
switch(input.size() % 3)
{
case 1:
temp = (*it++) << 16;
encoded.append(1, kEncodeLookup[(temp & 0x00FC0000) >> 18]);
encoded.append(1, kEncodeLookup[(temp & 0x0003F000) >> 12]);
encoded.append(2, kPadCharacter);
break;
case 2:
temp = (*it++) << 16;
temp += (*it++) << 8;
encoded.append(1, kEncodeLookup[(temp & 0x00FC0000) >> 18]);
encoded.append(1, kEncodeLookup[(temp & 0x0003F000) >> 12]);
encoded.append(1, kEncodeLookup[(temp & 0x00000FC0) >> 6 ]);
encoded.append(1, kPadCharacter);
break;
}
return encoded;
}
std::vector<byte> decode(const std::string& input)
{
if(input.length() % 4)
throw std::runtime_error("Invalid base64 length!");
std::size_t padding{};
if(input.length())
{
if(input[input.length() - 1] == kPadCharacter) padding++;
if(input[input.length() - 2] == kPadCharacter) padding++;
}
std::vector<byte> decoded;
decoded.reserve(((input.length() / 4) * 3) - padding);
std::uint32_t temp{};
auto it = input.begin();
while(it < input.end())
{
for(std::size_t i = 0; i < 4; ++i)
{
temp <<= 6;
if (*it >= 0x41 && *it <= 0x5A) temp |= *it - 0x41;
else if(*it >= 0x61 && *it <= 0x7A) temp |= *it - 0x47;
else if(*it >= 0x30 && *it <= 0x39) temp |= *it + 0x04;
else if(*it == 0x2B) temp |= 0x3E;
else if(*it == 0x2F) temp |= 0x3F;
else if(*it == kPadCharacter)
{
switch(input.end() - it)
{
case 1:
decoded.push_back((temp >> 16) & 0x000000FF);
decoded.push_back((temp >> 8 ) & 0x000000FF);
return decoded;
case 2:
decoded.push_back((temp >> 10) & 0x000000FF);
return decoded;
default:
throw std::runtime_error("Invalid padding in base64!");
}
}
else throw std::runtime_error("Invalid character in base64!");
++it;
}
decoded.push_back((temp >> 16) & 0x000000FF);
decoded.push_back((temp >> 8 ) & 0x000000FF);
decoded.push_back((temp ) & 0x000000FF);
}
return decoded;
}
}
答案 12 :(得分：0)
这是我写的一个，它使用联合和位域来最大化效率和可读性。
const char PADDING_CHAR = '=';
const char* ALPHABET = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
const uint8_t DECODED_ALPHBET[128]={0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,62,0,0,0,63,52,53,54,55,56,57,58,59,60,61,0,0,0,0,0,0,0,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,0,0,0,0,0,0,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,0,0,0,0,0};
/**
* Given a string, this function will encode it in 64b (with padding)
*/
std::string encodeBase64(const std::string& binaryText)
{
std::string encoded((binaryText.size()/3 + (binaryText.size()%3 > 0)) << 2, PADDING_CHAR);
const char* bytes = binaryText.data();
union
{
uint32_t temp = 0;
struct
{
uint32_t first : 6, second : 6, third : 6, fourth : 6;
} tempBytes;
};
std::string::iterator currEncoding = encoded.begin();
for(uint32_t i = 0, lim = binaryText.size() / 3; i < lim; ++i, bytes+=3)
{
temp = bytes[0] << 16 | bytes[1] << 8 | bytes[2];
(*currEncoding++) = ALPHABET[tempBytes.fourth];
(*currEncoding++) = ALPHABET[tempBytes.third];
(*currEncoding++) = ALPHABET[tempBytes.second];
(*currEncoding++) = ALPHABET[tempBytes.first];
}
switch(binaryText.size() % 3)
{
case 1:
temp = bytes[0] << 16;
(*currEncoding++) = ALPHABET[tempBytes.fourth];
(*currEncoding++) = ALPHABET[tempBytes.third];
break;
case 2:
temp = bytes[0] << 16 | bytes[1] << 8;
(*currEncoding++) = ALPHABET[tempBytes.fourth];
(*currEncoding++) = ALPHABET[tempBytes.third];
(*currEncoding++) = ALPHABET[tempBytes.second];
break;
}
return encoded;
}
/**
* Given a 64b padding-encoded string, this function will decode it.
*/
std::string decodeBase64(const std::string& base64Text)
{
if( base64Text.empty() )
return "";
assert((base64Text.size()&3) == 0 && "The base64 text to be decoded must have a length devisible by 4!");
uint32_t numPadding = (*std::prev(base64Text.end(),1) == PADDING_CHAR) + (*std::prev(base64Text.end(),2) == PADDING_CHAR);
std::string decoded((base64Text.size()*3>>2) - numPadding, '.');
union
{
uint32_t temp;
char tempBytes[4];
};
const uint8_t* bytes = reinterpret_cast<const uint8_t*>(base64Text.data());
std::string::iterator currDecoding = decoded.begin();
for(uint32_t i = 0, lim = (base64Text.size() >> 2) - (numPadding!=0); i < lim; ++i, bytes+=4)
{
temp = DECODED_ALPHBET[bytes[0]] << 18 | DECODED_ALPHBET[bytes[1]] << 12 | DECODED_ALPHBET[bytes[2]] << 6 | DECODED_ALPHBET[bytes[3]];
(*currDecoding++) = tempBytes[2];
(*currDecoding++) = tempBytes[1];
(*currDecoding++) = tempBytes[0];
}
switch (numPadding)
{
case 2:
temp = DECODED_ALPHBET[bytes[0]] << 18 | DECODED_ALPHBET[bytes[1]] << 12;
(*currDecoding++) = tempBytes[2];
break;
case 1:
temp = DECODED_ALPHBET[bytes[0]] << 18 | DECODED_ALPHBET[bytes[1]] << 12 | DECODED_ALPHBET[bytes[2]] << 6;
(*currDecoding++) = tempBytes[2];
(*currDecoding++) = tempBytes[1];
break;
}
return decoded;
}