给定一个十六进制值的字符串,例如 "0011223344",其中每两个字符代表一个十六进制值,如 0x00、0x11 等等。
我该如何将这些值添加到 char 数组中?
相当于说:
char array[4] = { 0x00, 0x11 ... };
你不能将 5 字节的数据存入一个 4 字节的数组中;这会导致缓冲区溢出。
如果你有一个十六进制数字的字符串,可以使用 sscanf()
和一个循环:
#include <stdio.h>
#include <ctype.h>
int main()
{
const char *src = "0011223344";
char buffer[5];
char *dst = buffer;
char *end = buffer + sizeof(buffer);
unsigned int u;
while (dst < end && sscanf(src, "%2x", &u) == 1)
{
*dst++ = u;
src += 2;
}
for (dst = buffer; dst < end; dst++)
printf("%d: %c (%d, 0x%02x)\n", dst - buffer,
(isprint(*dst) ? *dst : '.'), *dst, *dst);
return(0);
}
注意打印以零字节开头的字符串需要小心处理;大多数操作会在遇到第一个空字节时终止。请注意,此代码未对缓冲区进行空字符终止;不清楚是否需要空字符终止,并且我声明的缓冲区没有足够的空间添加终止符号(但很容易修复)。如果将代码打包为子程序,则有很大机会需要返回转换后字符串的长度(尽管您也可以认为它是源字符串长度除以二的长度)。
#define hex(c) ((*(c)>='a')?*(c)-'a'+10:(*(c)>='A')?*(c)-'A'+10:*(c)-'0')
void hex2char( char *to ){
for(char *from=to; *from; from+=2) *to++=hex(from)*16+hex(from+1);
*to=0;
}
编辑1:抱歉,我忘记计算字母A-F(a-f)的值。
编辑2:我尝试编写更加严谨的代码:
#include <string.h>
int xdigit( char digit ){
int val;
if( '0' <= digit && digit <= '9' ) val = digit -'0';
else if( 'a' <= digit && digit <= 'f' ) val = digit -'a'+10;
else if( 'A' <= digit && digit <= 'F' ) val = digit -'A'+10;
else val = -1;
return val;
}
int xstr2str( char *buf, unsigned bufsize, const char *in ){
if( !in ) return -1; // missing input string
unsigned inlen=strlen(in);
if( inlen%2 != 0 ) return -2; // hex string must even sized
for( unsigned i=0; i<inlen; i++ )
if( xdigit(in[i])<0 ) return -3; // bad character in hex string
if( !buf || bufsize<inlen/2+1 ) return -4; // no buffer or too small
for( unsigned i=0,j=0; i<inlen; i+=2,j++ )
buf[j] = xdigit(in[i])*16 + xdigit(in[i+1]);
buf[inlen/2] = '\0';
return inlen/2+1;
}
测试:
#include <stdio.h>
char buf[100] = "test";
void test( char *buf, const char *s ){
printf("%3i=xstr2str( \"%s\", 100, \"%s\" )\n", xstr2str( buf, 100, s ), buf, s );
}
int main(){
test( buf, (char*)0 );
test( buf, "123" );
test( buf, "3x" );
test( (char*)0, "" );
test( buf, "" );
test( buf, "3C3e" );
test( buf, "3c31323e" );
strcpy( buf, "616263" ); test( buf, buf );
}
结果:
-1=xstr2str( "test", 100, "(null)" )
-2=xstr2str( "test", 100, "123" )
-3=xstr2str( "test", 100, "3x" )
-4=xstr2str( "(null)", 100, "" )
1=xstr2str( "", 100, "" )
3=xstr2str( "", 100, "3C3e" )
5=xstr2str( "", 100, "3c31323e" )
4=xstr2str( "abc", 100, "abc" )
char array[] = {0, 17, 34, 51, 68};
。但我认为提问者在说"i.e."时实际上是指"e.g."。 - Steve Jessop// Convert from ascii hex representation to binary
// Examples;
// "00" -> 0
// "2a" -> 42
// "ff" -> 255
// Case insensitive, 2 characters of input required, no error checking
int hex2bin( const char *s )
{
int ret=0;
int i;
for( i=0; i<2; i++ )
{
char c = *s++;
int n=0;
if( '0'<=c && c<='9' )
n = c-'0';
else if( 'a'<=c && c<='f' )
n = 10 + c-'a';
else if( 'A'<=c && c<='F' )
n = 10 + c-'A';
ret = n + ret*16;
}
return ret;
}
int main()
{
const char *in = "0011223344";
char out[5];
int i;
// Hex to binary conversion loop. For example;
// If in="0011223344" set out[] to {0x00,0x11,0x22,0x33,0x44}
for( i=0; i<5; i++ )
{
out[i] = hex2bin( in );
in += 2;
}
return 0;
}
/* not my original work, on stacko somewhere ? */
for (i=0;i < 4;i++) {
char a = string[2 * i];
char b = string[2 * i + 1];
array[i] = (((encode(a) * 16) & 0xF0) + (encode(b) & 0x0F));
}
并且函数encode()被定义...
unsigned char encode(char x) { /* Function to encode a hex character */
/****************************************************************************
* these offsets should all be decimal ..x validated for hex.. *
****************************************************************************/
if (x >= '0' && x <= '9') /* 0-9 is offset by hex 30 */
return (x - 0x30);
else if (x >= 'a' && x <= 'f') /* a-f offset by hex 57 */
return(x - 0x57);
else if (x >= 'A' && x <= 'F') /* A-F offset by hex 37 */
return(x - 0x37);
}
hex(0x30,0x57,0x37)
这个提取过程的吗? - Faruk// in = "63 09 58 81"
void hexatoascii(char *in, char* out, int len){
char buf[5000];
int i,j=0;
char * data[5000];
printf("\n size %d", strlen(in));
for (i = 0; i < strlen(in); i+=2)
{
data[j] = (char*)malloc(8);
if (in[i] == ' '){
i++;
}
else if(in[i + 1] == ' '){
i++;
}
printf("\n %c%c", in[i],in[i+1]);
sprintf(data[j], "%c%c", in[i], in[i+1]);
j++;
}
for (i = 0; i < j-1; i++){
int tmp;
printf("\n data %s", data[i] );
sscanf(data[i], "%2x", &tmp);
out[i] = tmp;
}
//printf("\n ascii value of hexa %s", out);
}
int hex2bin_by_zibri(char *source_str, char *dest_buffer)
{
char *line = source_str;
char *data = line;
int offset;
int read_byte;
int data_len = 0;
while (sscanf(data, " %02x%n", &read_byte, &offset) == 1) {
dest_buffer[data_len++] = read_byte;
data += offset;
}
return data_len;
}
提供最佳方法:
将十六进制字符串转换为数值,例如 str[] = "0011223344" 转换为值 0x0011223344,使用以下方法:
value = strtoul(string, NULL, 16); // or strtoull()
完成。如果需要删除开头的0x00,请参见下文。
对于LITTLE_ENDIAN平台,还需注意以下内容: 将十六进制值转换为字符数组,例如将值0x11223344转换为char arr[N] = {0x00, 0x11, ...}
unsigned long *hex = (unsigned long*)arr;
*hex = htonl(value);
// you'd like to remove any beginning 0x00
char *zero = arr;
while (0x00 == *zero) { zero++; }
if (zero > arr) memmove(zero, arr, sizeof(arr) - (zero - arr));
完成。
注: 在32位系统上将长字符串转换为64位十六进制字符数组时,应使用unsigned long long而不是unsigned long,并且htonl不足以完成此操作,因此请按照以下方式自行完成,因为可能没有htonll、htonq或hton64等函数:
#if __KERNEL__
/* Linux Kernel space */
#if defined(__LITTLE_ENDIAN_BITFIELD)
#define hton64(x) __swab64(x)
#else
#define hton64(x) (x)
#endif
#elif defined(__GNUC__)
/* GNU, user space */
#if __BYTE_ORDER == __LITTLE_ENDIAN
#define hton64(x) __bswap_64(x)
#else
#define hton64(x) (x)
#endif
#elif
...
#endif
#define ntoh64(x) hton64(x)
看一下http://effocore.googlecode.com/svn/trunk/devel/effo/codebase/builtin/include/impl/sys/bswap.h
致命错误...
有几种方法可以做到这一点...首先,您可以使用memcpy()将精确表示复制到char数组中。
您也可以使用位移和位掩码技术。我猜这就是你需要做的,因为它听起来像是一个作业问题。
最后,您可以使用一些花哨的指针间接引用来复制所需的内存位置。
所有这些方法都在这里详细说明:
{
char szVal[] = "268484927472";
char szOutput[30];
size_t nLen = strlen(szVal);
// Make sure it is even.
if ((nLen % 2) == 1)
{
printf("Error string must be even number of digits %s", szVal);
}
// Process each set of characters as a single character.
nLen >>= 1;
for (size_t idx = 0; idx < nLen; idx++)
{
char acTmp[3];
sscanf(szVal + (idx << 1), "%2s", acTmp);
szOutput[idx] = (char)strtol(acTmp, NULL, 16);
}
}
hex2bin
和bin2hex
实现。-1
表示无效的十六进制字符串)static char h2b(char c) {
return '0'<=c && c<='9' ? c - '0' :
'A'<=c && c<='F' ? c - 'A' + 10 :
'a'<=c && c<='f' ? c - 'a' + 10 :
/* else */ -1;
}
int hex2bin(unsigned char* bin, unsigned int bin_len, const char* hex) {
for(unsigned int i=0; i<bin_len; i++) {
char b[2] = {h2b(hex[2*i+0]), h2b(hex[2*i+1])};
if(b[0]<0 || b[1]<0) return -1;
bin[i] = b[0]*16 + b[1];
}
return 0;
}
static char b2h(unsigned char b, int upper) {
return b<10 ? '0'+b : (upper?'A':'a')+b-10;
}
void bin2hex(char* hex, const unsigned char* bin, unsigned int bin_len, int upper) {
for(unsigned int i=0; i<bin_len; i++) {
hex[2*i+0] = b2h(bin[i]>>4, upper);
hex[2*i+1] = b2h(bin[i]&0x0F, upper);
}
}