C语言实现 Base64 和 Base32 编解码

BrightXu

C语言实现 Base64 和 Base32 编解码

Base64 编解码

定义头文件 base64.h

#ifndef _BASE64_H
#define _BASE64_H

#include <stdint.h>

static const char BASE64_MAP[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=";
static const uint8_t BASE64_REVERSE_MAP[] = {
        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,
};

char *base64encode(const char *str, uint64_t len);

char *base64parse(const char *base64Str, uint64_t len);

#endif //_BASE64_H

思路

Base64编码就是用64个可打印字符表示二进制数据。

$$64 = 2^6$$

，所以需要6 Bit来表示一个base64字符。一个字节8 Bit，6和8的最小公倍数是24。编码的过程中，以3个字节为一组转为4个base64字符，不足3个字节以0代替。为方便转换，以一个无符号32位整数(uint32_t)为中间载体。先由高位到低位将这三个字节填充到这个整数中，然后由高位到低位依次读取6位，获取对应数值的字母，共读取4次。如下图所示。解码的过程是上述的逆过程。

实现

#include "base64.h"
#include <stdlib.h>

#ifndef CEIL_POS
// 正数向上取整 CEIL_POS(2.345) => 3
#define CEIL_POS(X) (X > (uint64_t)(X) ? (uint64_t)(X+1) : (uint64_t)(X))
#endif

char *base64encode(const char *str, uint64_t len) {
    uint64_t length = CEIL_POS(len * 4 / 3) + 1;
    char *base64Chars = (char *) malloc(sizeof(char) * length);
    uint64_t idx = 0;

    for (uint64_t i = 0; i < len; i += 3) {
        uint32_t byte1 = (uint8_t) str[i];
        uint16_t byte2 = (i + 1 < len) ? (uint8_t) str[i + 1] : 0;
        uint8_t byte3 = (i + 2 < len) ? (uint8_t) str[i + 2] : 0;

        uint32_t triplet = (byte1 << 16) | (byte2 << 8) | byte3;

        for (uint64_t j = 0; (j < 4) && (i + j * 0.75 < len); j++) {
            base64Chars[idx] = BASE64_MAP[(triplet >> (6 * (3 - j))) & 0x3f];
            idx++;
        }
    }

    char paddingChar = BASE64_MAP[64];
    if (paddingChar) {
        while (idx % 4) {
            base64Chars[idx] = paddingChar;
            idx++;
        }
    }
    base64Chars[idx] = 0;
    return base64Chars;
}

char *base64parse(const char *base64Str, uint64_t len) {
    while (base64Str[len - 1] == BASE64_MAP[64]) {
        len--;
    }
    uint64_t length = CEIL_POS(len * 3 / 4) + 1;
    char *str = (char *) malloc(sizeof(char) * length);
    uint64_t idx = 0;

    for (uint64_t i = 0; i < len; i += 4) {
        uint32_t triplet = 0;
        for (uint8_t j = 0; j < 4; ++j) {
            if (i + j < len) triplet = (triplet << 6) | ((uint8_t) BASE64_REVERSE_MAP[base64Str[i + j]] & 0x3f);
            else triplet = triplet << 6;
        }
        for (uint8_t j = 0; (j < 3); ++j) {
            str[idx] = (triplet >> (8 * (2 - j))) & 0xff;
            idx++;
        }
    }
    str[idx] = 0;
    return str;
}

使用

// test_b64.c
#include <stdio.h>
#include <string.h>
#include "base64.h"

int main() {
    char str1[] = "Man is distinguished, not only by his reason, but by this singular passion from other animals, which is a lust of the mind, that by a perseverance of delight in the continued and indefatigable generation of knowledge, exceeds the short vehemence of any carnal pleasure.";
    char *encoded = base64encode(str1, strlen(str1));
    puts(encoded);

    char str2[] = "TWFuIGlzIGRpc3Rpbmd1aXNoZWQsIG5vdCBvbmx5IGJ5IGhpcyByZWFzb24sIGJ1dCBieSB0aGlzIHNpbmd1bGFyIHBhc3Npb24gZnJvbSBvdGhlciBhbmltYWxzLCB3aGljaCBpcyBhIGx1c3Qgb2YgdGhlIG1pbmQsIHRoYXQgYnkgYSBwZXJzZXZlcmFuY2Ugb2YgZGVsaWdodCBpbiB0aGUgY29udGludWVkIGFuZCBpbmRlZmF0aWdhYmxlIGdlbmVyYXRpb24gb2Yga25vd2xlZGdlLCBleGNlZWRzIHRoZSBzaG9ydCB2ZWhlbWVuY2Ugb2YgYW55IGNhcm5hbCBwbGVhc3VyZS4=";
    char *decoded = base64parse(str2, strlen(str2));
    puts(decoded);
    return 0;
}

Base32 编解码

定义头文件 base32.h

#ifndef _BASE32_H
#define _BASE32_H

#include <stdint.h>

static const char BASE32_MAP[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ234567=";
static const uint8_t BASE32_REVERSE_MAP[] = {
        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, 0, 0, 0, 0, 0, 0, 0, 26, 27, 28, 29, 30, 31, 0, 0, 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,
};

char *base32encode(const char *str, uint64_t len);

char *base32parse(const char *base32Str, uint64_t len);

#endif //_BASE32_H

思路

Base32编码就是用32个可打印字符表示二进制数据。

$$32 = 2^5$$

，所以需要5 Bit来表示一个base32字符。一个字节8 Bit，5和8的最小公倍数是40。编码的过程中，以5个字节为一组转为8个base32字符，不足5个字节以0代替。为方便转换，以一个无符号64位整数(uint64_t)为中间载体。先由高位到低位将这5个字节填充到这个整数中，然后由高位到低位依次读取5位，获取对应数值的字母，共读取8次。如下图所示。解码的过程是上述的逆过程。

实现

#include "base32.h"
#include <stdlib.h>

#ifndef CEIL_POS
#define CEIL_POS(X) (X > (uint64_t)(X) ? (uint64_t)(X+1) : (uint64_t)(X))
#endif

char *base32encode(const char *str, uint64_t len) {
    uint64_t length = CEIL_POS(len * 8 / 5) + 1;
    char *base32Chars = (char *) malloc(sizeof(char) * length);
    uint64_t idx = 0;

    for (uint64_t i = 0; i < len; i += 5) {
        uint64_t byte1 = (uint8_t) str[i];
        uint64_t byte2 = (i + 1 < len) ? (uint8_t) str[i + 1] : 0;
        uint32_t byte3 = (i + 2 < len) ? (uint8_t) str[i + 2] : 0;
        uint16_t byte4 = (i + 3 < len) ? (uint8_t) str[i + 3] : 0;
        uint8_t byte5 = (i + 4 < len) ? (uint8_t) str[i + 4] : 0;

        uint64_t quintuple = (byte1 << 32) | (byte2 << 24) | (byte3 << 16) | (byte4 << 8) | byte5;

        for (uint64_t j = 0; (j < 8) && (i + j * 0.625 < len); j++) {
            base32Chars[idx] = BASE32_MAP[(quintuple >> (5 * (7 - j))) & 0x1f];
            idx++;
        }
    }

    char paddingChar = BASE32_MAP[32];
    if (paddingChar) {
        while (idx % 8) {
            base32Chars[idx] = paddingChar;
            idx++;
        }
    }
    base32Chars[idx] = 0;
    return base32Chars;
}

char *base32parse(const char *base32Str, uint64_t len) {
    while (base32Str[len - 1] == BASE32_MAP[32]) {
        len--;
    }
    uint64_t length = CEIL_POS(len * 5 / 8) + 1;
    char *str = (char *) malloc(sizeof(char) * length);
    uint64_t idx = 0;

    for (uint64_t i = 0; i < len; i += 8) {
        uint64_t quintuple = 0;
        for (uint8_t j = 0; j < 8; ++j) {
            if (i + j < len) quintuple = (quintuple << 5) | ((uint8_t) BASE32_REVERSE_MAP[base32Str[i + j]] & 0x1f);
            else quintuple = quintuple << 5;
        }
        for (uint8_t j = 0; (j < 5); ++j) {
            str[idx] = (quintuple >> (8 * (4 - j))) & 0xff;
            idx++;
        }
    }
    str[idx] = 0;
    return str;
}

使用

// test_b32.c
#include <stdio.h>
#include <string.h>
#include "base32.h"

int main() {
    char str1[] = "Man is distinguished, not only by his reason, but by this singular passion from other animals, which is a lust of the mind, that by a perseverance of delight in the continued and indefatigable generation of knowledge, exceeds the short vehemence of any carnal pleasure.";
    char *encoded = base32encode(str1, strlen(str1));
    puts(encoded);

    char str2[] = "JVQW4IDJOMQGI2LTORUW4Z3VNFZWQZLEFQQG433UEBXW43DZEBRHSIDINFZSA4TFMFZW63RMEBRHK5BAMJ4SA5DINFZSA43JNZTXK3DBOIQHAYLTONUW63RAMZZG63JAN52GQZLSEBQW42LNMFWHGLBAO5UGSY3IEBUXGIDBEBWHK43UEBXWMIDUNBSSA3LJNZSCYIDUNBQXIIDCPEQGCIDQMVZHGZLWMVZGC3TDMUQG6ZRAMRSWY2LHNB2CA2LOEB2GQZJAMNXW45DJNZ2WKZBAMFXGIIDJNZSGKZTBORUWOYLCNRSSAZ3FNZSXEYLUNFXW4IDPMYQGW3TPO5WGKZDHMUWCAZLYMNSWKZDTEB2GQZJAONUG64TUEB3GK2DFNVSW4Y3FEBXWMIDBNZ4SAY3BOJXGC3BAOBWGKYLTOVZGKLQ=";
    char *decoded = base32parse(str2, strlen(str2));
    puts(decoded);
    return 0;
}

BrightXu

zhukoov 发表于 2022-11-20 13:47
楼主的代码给了我很大的启发.得知Base64的译码表即破解.那么在传输过程中如何准确识别Base64呢?

这跟破解有什么关系？base64只是一种数据编码方式，不存在加密解密之说，又何来破解可言？就像十六进制和十进制之间的关系，只是一种数据表示形式而已，base64编码后的内容只是对人类来说不可直接看懂罢了。而且Base64码表也有统一的规范和标准，大多数情况用的都是通用的码表。

zhukoov

BrightXu 发表于 2022-11-21 01:20
这跟破解有什么关系？base64只是一种数据编码方式，不存在加密解密之说，又何来破解可言？就像十六进制和 ...

原来如此，谢谢楼主纠正，我还以为它本身是一种加密呢。现在我了解了，非常感谢。

sknbs

学习了，这就去试一下

QS69

这都能行？

blindcat

膜拜大佬

l441669899

感谢楼主！

MI20220721

有些看不懂，先学习学习！

lxfw2000

学习了。下来找机会练一下手！

怜渠客

自实现函数，学习学习

metoo2

感谢楼主~~