C语言 十六进制字符串到C中的字节数组
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/3408706/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Hexadecimal string to byte array in C
提问by Szere Dyeri
Is there any standard C function that converts from hexadecimal string to byte array?
I do not want to write my own function.
是否有任何标准 C 函数可以将十六进制字符串转换为字节数组？
我不想编写自己的函数。
回答by Michael Foukarakis
As far as I know, there's no standard function to do so, but it's simple to achieve in the following manner:
据我所知，没有标准功能可以这样做，但通过以下方式实现很简单：
#include <stdio.h>
int main(int argc, char **argv) {
const char hexstring[] = "DEadbeef10203040b00b1e50", *pos = hexstring;
unsigned char val[12];
/* WARNING: no sanitization or error-checking whatsoever */
for (size_t count = 0; count < sizeof val/sizeof *val; count++) {
sscanf(pos, "%2hhx", &val[count]);
pos += 2;
}
printf("0x");
for(size_t count = 0; count < sizeof val/sizeof *val; count++)
printf("%02x", val[count]);
printf("\n");
return 0;
}
Edit
编辑
As Al pointed out, in case of an odd number of hex digits in the string, you have to make sure you prefix it with a starting 0. For example, the string "f00f5"will be evaluated as {0xf0, 0x0f, 0x05}erroneously by the above example, instead of the proper {0x0f, 0x00, 0xf5}.
正如 Al 指出的那样，如果字符串中有奇数个十六进制数字，您必须确保以 0 开头。例如， 上面的示例"f00f5"将{0xf0, 0x0f, 0x05}错误地评估字符串，而不是正确的{0x0f, 0x00, 0xf5}.
Amended the example a little bit to address the comment from @MassimoCallegari
稍微修改了示例以解决@MassimoCallegari 的评论
回答by David M. Syzdek
I found this question by Googling for the same thing. I don't like the idea of calling sscanf() or strtol() since it feels like overkill. I wrote a quick function which does not validate that the text is indeed the hexadecimal presentation of a byte stream, but will handle odd number of hex digits:
我通过谷歌搜索发现了同样的问题。我不喜欢调用 sscanf() 或 strtol() 的想法，因为这感觉有点矫枉过正。我写了一个快速函数，它不会验证文本确实是字节流的十六进制表示，但会处理奇数个十六进制数字：
uint8_t tallymarker_hextobin(const char * str, uint8_t * bytes, size_t blen)
{
uint8_t pos;
uint8_t idx0;
uint8_t idx1;
// mapping of ASCII characters to hex values
const uint8_t hashmap[] =
{
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // !"#$%&'
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ()*+,-./
0x00, 0x01, 0x02, 0x03, 0x04, 0x05, 0x06, 0x07, // 01234567
0x08, 0x09, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // 89:;<=>?
0x00, 0x0a, 0x0b, 0x0c, 0x0d, 0x0e, 0x0f, 0x00, // @ABCDEFG
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // HIJKLMNO
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // PQRSTUVW
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // XYZ[\]^_
0x00, 0x0a, 0x0b, 0x0c, 0x0d, 0x0e, 0x0f, 0x00, // `abcdefg
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // hijklmno
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // pqrstuvw
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // xyz{|}~.
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, // ........
0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00 // ........
};
bzero(bytes, blen);
for (pos = 0; ((pos < (blen*2)) && (pos < strlen(str))); pos += 2)
{
idx0 = (uint8_t)str[pos+0];
idx1 = (uint8_t)str[pos+1];
bytes[pos/2] = (uint8_t)(hashmap[idx0] << 4) | hashmap[idx1];
};
return(0);
}
回答by Mike M
Apart from the excellent answers above I though I would write a C function that does not use any libraries and has some guards against bad strings.
除了上面的优秀答案之外，我虽然会编写一个不使用任何库并且对坏字符串有一些保护措施的 C 函数。
uint8_t* datahex(char* string) {
if(string == NULL)
return NULL;
size_t slength = strlen(string);
if((slength % 2) != 0) // must be even
return NULL;
size_t dlength = slength / 2;
uint8_t* data = malloc(dlength);
memset(data, 0, dlength);
size_t index = 0;
while (index < slength) {
char c = string[index];
int value = 0;
if(c >= '0' && c <= '9')
value = (c - '0');
else if (c >= 'A' && c <= 'F')
value = (10 + (c - 'A'));
else if (c >= 'a' && c <= 'f')
value = (10 + (c - 'a'));
else {
free(data);
return NULL;
}
data[(index/2)] += value << (((index + 1) % 2) * 4);
index++;
}
return data;
}
Explanation:
解释：
a. index / 2| Division between integers will round down the value, so 0/2 = 0, 1/2 = 0, 2/2 = 1, 3/2 = 1, 4/2 = 2, 5/2 = 2, etc. So, for every 2 string characters we add the value to 1 data byte.
一种。指数 / 2| 整数之间的除法将向下取整值，因此 0/2 = 0、1/2 = 0、2/2 = 1、3/2 = 1、4/2 = 2、5/2 = 2 等。所以，对于每 2 个字符串字符，我们将值添加到 1 个数据字节。
b. (index + 1) % 2| We want odd numbers to result to 1 and even to 0 since the first digit of a hex string is the most significant and needs to be multiplied by 16. so for index 0 => 0 + 1 % 2 = 1, index 1 => 1 + 1 % 2 = 0 etc.
湾 (指数 + 1) % 2| 我们希望奇数为 1，偶数为 0，因为十六进制字符串的第一个数字是最重要的，需要乘以 16。所以对于索引 0 => 0 + 1 % 2 = 1，索引 1 => 1 + 1 % 2 = 0 等等。
c. << 4| Shift by 4 is multiplying by 16. example: b00000001 << 4 = b00010000
C。<< 4| 移位 4 就是乘以 16。例如：b00000001 << 4 = b00010000
回答by dmckee --- ex-moderator kitten
For short strings, strtol, strtoll, and strtoimaxwill work just fine (note that the third argument is the base to use in processing the string...set it to 16). If your input is longer than number-of-bits-in-the-longest-integer-type/4then you'll need one of the more flexible methods suggested by other answers.
对于短字符串，strtol, strtoll, 和strtoimax将工作得很好（请注意，第三个参数是用于处理字符串的基础...将其设置为 16）。如果您的输入比number-of-bits-in-the-longest-integer-type/4那时长，您将需要其他答案建议的更灵活的方法之一。
回答by mgalela
By some modification form user411313's code, following works for me:
通过对 user411313 代码的一些修改，以下对我有用：
#include <stdio.h>
#include <stdint.h>
#include <string.h>
int main ()
{
char *hexstring = "deadbeef10203040b00b1e50";
int i;
unsigned int bytearray[12];
uint8_t str_len = strlen(hexstring);
for (i = 0; i < (str_len / 2); i++) {
sscanf(hexstring + 2*i, "%02x", &bytearray[i]);
printf("bytearray %d: %02x\n", i, bytearray[i]);
}
return 0;
}
回答by SMH
A fleshed out version of Michael Foukarakis post (since I don't have the "reputation" to add a comment to that post yet):
Michael Fokarakis 帖子的充实版本（因为我还没有“声誉”来为该帖子添加评论）：
#include <stdio.h>
#include <string.h>
void print(unsigned char *byte_array, int byte_array_size)
{
int i = 0;
printf("0x");
for(; i < byte_array_size; i++)
{
printf("%02x", byte_array[i]);
}
printf("\n");
}
int convert(const char *hex_str, unsigned char *byte_array, int byte_array_max)
{
int hex_str_len = strlen(hex_str);
int i = 0, j = 0;
// The output array size is half the hex_str length (rounded up)
int byte_array_size = (hex_str_len+1)/2;
if (byte_array_size > byte_array_max)
{
// Too big for the output array
return -1;
}
if (hex_str_len % 2 == 1)
{
// hex_str is an odd length, so assume an implicit "0" prefix
if (sscanf(&(hex_str[0]), "%1hhx", &(byte_array[0])) != 1)
{
return -1;
}
i = j = 1;
}
for (; i < hex_str_len; i+=2, j++)
{
if (sscanf(&(hex_str[i]), "%2hhx", &(byte_array[j])) != 1)
{
return -1;
}
}
return byte_array_size;
}
void main()
{
char *examples[] = { "", "5", "D", "5D", "5Df", "deadbeef10203040b00b1e50", "02invalid55" };
unsigned char byte_array[128];
int i = 0;
for (; i < sizeof(examples)/sizeof(char *); i++)
{
int size = convert(examples[i], byte_array, 128);
if (size < 0)
{
printf("Failed to convert '%s'\n", examples[i]);
}
else if (size == 0)
{
printf("Nothing to convert for '%s'\n", examples[i]);
}
else
{
print(byte_array, size);
}
}
}
回答by flcoder
Following is the solution I wrote up for performance reasons:
以下是我出于性能原因编写的解决方案：
void hex2bin(const char* in, size_t len, unsigned char* out) {
static const unsigned char TBL[] = {
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 58, 59, 60, 61,
62, 63, 64, 10, 11, 12, 13, 14, 15, 71, 72, 73, 74, 75,
76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89,
90, 91, 92, 93, 94, 95, 96, 10, 11, 12, 13, 14, 15
};
static const unsigned char *LOOKUP = TBL - 48;
const char* end = in + len;
while(in < end) *(out++) = LOOKUP[*(in++)] << 4 | LOOKUP[*(in++)];
}
Example:
例子：
unsigned char seckey[32];
hex2bin("351aaaec0070d13d350afae2bc43b68c7e590268889869dde489f2f7988f3fee", 64, seckey);
/*
seckey = {
53, 26, 170, 236, 0, 112, 209, 61, 53, 10, 250, 226, 188, 67, 182, 140,
126, 89, 2, 104, 136, 152, 105, 221, 228, 137, 242, 247, 152, 143, 63, 238
};
*/
If you don't need to support lowercase:
如果您不需要支持小写：
static const unsigned char TBL[] = {
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 58, 59,
60, 61, 62, 63, 64, 10, 11, 12, 13, 14, 15
};
回答by Henri Socha
Here is HexToBin and BinToHex relatively clean and readable. (Note originally there were returned enum error codes through an error logging system not a simple -1 or -2.)
这里是 HexToBin 和 BinToHex 相对干净和可读。（请注意，最初通过错误记录系统返回了枚举错误代码，而不是简单的 -1 或 -2。）
typedef unsigned char ByteData;
ByteData HexChar (char c)
{
if ('0' <= c && c <= '9') return (ByteData)(c - '0');
if ('A' <= c && c <= 'F') return (ByteData)(c - 'A' + 10);
if ('a' <= c && c <= 'f') return (ByteData)(c - 'a' + 10);
return (ByteData)(-1);
}
ssize_t HexToBin (const char* s, ByteData * buff, ssize_t length)
{
ssize_t result = 0;
if (!s || !buff || length <= 0) return -2;
while (*s)
{
ByteData nib1 = HexChar(*s++);
if ((signed)nib1 < 0) return -3;
ByteData nib2 = HexChar(*s++);
if ((signed)nib2 < 0) return -4;
ByteData bin = (nib1 << 4) + nib2;
if (length-- <= 0) return -5;
*buff++ = bin;
++result;
}
return result;
}
void BinToHex (const ByteData * buff, ssize_t length, char * output, ssize_t outLength)
{
char binHex[] = "0123456789ABCDEF";
if (!output || outLength < 4) return (void)(-6);
*output = '#ifndef HEX_TOOLS_H
#define HEX_TOOLS_H
char *bin2hex(unsigned char*, int);
unsigned char *hex2bin(const char*);
#endif // HEX_TOOLS_H
';
if (!buff || length <= 0 || outLength <= 2 * length)
{
memcpy(output, "ERR", 4);
return (void)(-7);
}
for (; length > 0; --length, outLength -= 2)
{
ByteData byte = *buff++;
*output++ = binHex[(byte >> 4) & 0x0F];
*output++ = binHex[byte & 0x0F];
}
if (outLength-- <= 0) return (void)(-8);
*output++ = '#include <stdlib.h>
char *bin2hex(unsigned char *p, int len)
{
char *hex = malloc(((2*len) + 1));
char *r = hex;
while(len && p)
{
(*r) = ((*p) & 0xF0) >> 4;
(*r) = ((*r) <= 9 ? '0' + (*r) : 'A' - 10 + (*r));
r++;
(*r) = ((*p) & 0x0F);
(*r) = ((*r) <= 9 ? '0' + (*r) : 'A' - 10 + (*r));
r++;
p++;
len--;
}
*r = '#include <stdio.h>
#include "hextools.h"
int main(void)
{
unsigned char s[] = { 0xa0, 0xf9, 0xc3, 0xde, 0x44 };
char *hex = bin2hex(s, sizeof s);
puts(hex);
unsigned char *bin;
bin = hex2bin(hex);
puts(bin2hex(bin, 5));
size_t k;
for(k=0; k<5; k++)
printf("%02X", bin[k]);
putchar('\n');
return 0;
}
';
return hex;
}
unsigned char *hex2bin(const char *str)
{
int len, h;
unsigned char *result, *err, *p, c;
err = malloc(1);
*err = 0;
if (!str)
return err;
if (!*str)
return err;
len = 0;
p = (unsigned char*) str;
while (*p++)
len++;
result = malloc((len/2)+1);
h = !(len%2) * 4;
p = result;
*p = 0;
c = *str;
while(c)
{
if(('0' <= c) && (c <= '9'))
*p += (c - '0') << h;
else if(('A' <= c) && (c <= 'F'))
*p += (c - 'A' + 10) << h;
else if(('a' <= c) && (c <= 'f'))
*p += (c - 'a' + 10) << h;
else
return err;
str++;
c = *str;
if (h)
h = 0;
else
{
h = 4;
p++;
*p = 0;
}
}
return result;
}
';
}
回答by hutheano
hextools.h
hextools.h
In main()
{
printf("enter string :\n");
fgets(buf, 200, stdin);
unsigned char str_len = strlen(buf);
k=0;
unsigned char bytearray[100];
for(j=0;j<str_len-1;j=j+2)
{ bytearray[k++]=converttohex(&buffer[j]);
printf(" %02X",bytearray[k-1]);
}
}
Use this
int converttohex(char * val)
{
unsigned char temp = toupper(*val);
unsigned char fin=0;
if(temp>64)
temp=10+(temp-65);
else
temp=temp-48;
fin=(temp<<4)&0xf0;
temp = toupper(*(val+1));
if(temp>64)
temp=10+(temp-65);
else
temp=temp-48;
fin=fin|(temp & 0x0f);
return fin;
}
hextools.c
hextools.c
##代码##main.c
主文件
##代码##