我有一个包含十六进制值的文本文件。现在我需要将这些十六进制值转换为二进制并保存到另一个文件中。 但是我不知道如何将十六进制值转换为二进制字符串! 请帮忙...
我有一个包含十六进制值的文本文件。现在我需要将这些十六进制值转换为二进制并保存到另一个文件中。 但是我不知道如何将十六进制值转换为二进制字符串! 请帮忙...
非常简单,因为翻译是逐位进行的。
0 - 0000
1 - 0001
2 - 0010
3 - 0011
4 - 0100
5 - 0101
6 - 0110
7 - 0111
8 - 1000
9 - 1001
A - 1010
B - 1011
C - 1100
D - 1101
E - 1110
F - 1111
比如说,十六进制数FE2F8
在二进制里就是11111110001011111000
const char input[] = "..."; // the value to be converted
char res[9]; // the length of the output string has to be n+1 where n is the number of binary digits to show, in this case 8
res[8] = '\0';
int t = 128; // set this to s^(n-1) where n is the number of binary digits to show, in this case 8
int v = strtol(input, 0, 16); // convert the hex value to a number
while(t) // loop till we're done
{
strcat(res, t < v ? "1" : "0");
if(t < v)
v -= t;
t /= 2;
}
// res now contains the binary representation of the number
"0x3A"
这样的前缀):const char binary[16][5] = {"0000", "0001", "0010", "0011", "0100", ...};
const char digits = "0123456789abcdef";
const char input[] = "..." // input value
char res[1024];
res[0] = '\0';
int p = 0;
while(input[p])
{
const char *v = strchr(digits, tolower(input[p++]));
if (v)
strcat(res, binary[v - digits]);
}
// res now contains the binary representation of the number
strcat
根本没有填充res
数组,它始终保持为\0。 - scelong v = strtol({input[p++], 0}, NULL, 16); strcat(res, binary[v]);
。 - krb686$ { printf "char *testdata =\""; cat /dev/urandom \
| tr -d -c "0123456789abcdefABCDEF" \
| dd count=100 iflag=fullblock bs=1M; printf "\";\n" } > testdata.h
#!/usr/bin/env python
import sys,struct
sys.stdout.write("unsigned char base16_decoding_table1[256] = {\n")
for i in xrange(256):
try:
j = str(int(chr(i), 16))
except:
j = '0'
sys.stdout.write(j+',')
sys.stdout.write("};\n")
sys.stdout.write("\n")
l = 128*256*["0"]
for a in ['0','1','2','3','4','5','6','7','8','9','a','b','c','d','e','f','A','B','C','D','E','F']:
for b in ['0','1','2','3','4','5','6','7','8','9','a','b','c','d','e','f','A','B','C','D','E','F']:
l[struct.unpack("<H", a+b)[0]] = str(int(a+b, 16))
line = "unsigned char base16_decoding_table2[%d] = {"%(128*256)
for e in l:
line += e+","
if len(line) > 70:
sys.stdout.write(line+"\n")
line = ""
sys.stdout.write(line+"};\n")
sys.stdout.write("\n")
l = 256*256*["0"]
for a in ['0','1','2','3','4','5','6','7','8','9','a','b','c','d','e','f','A','B','C','D','E','F']:
for b in ['0','1','2','3','4','5','6','7','8','9','a','b','c','d','e','f','A','B','C','D','E','F']:
l[struct.unpack("<H", a+b)[0]] = str(int(a+b, 16))
line = "unsigned char base16_decoding_table3[%d] = {"%(256*256)
for e in l:
line += e+","
if len(line) > 70:
sys.stdout.write(line+"\n")
line = ""
sys.stdout.write(line+"};\n")
然后:
python gen.py > base16_decoding_table.h
现在我们可以编写一些 C 代码进行测试。
#include <stdio.h>
#include <time.h>
#include <inttypes.h>
#include "testdata.h"
#include "base16_decoding_table.h"
#define TESTDATALEN 104857600
/* the resulting binary string is half the size of the input hex string
* because every two hex characters map to one byte */
unsigned char result[TESTDATALEN/2];
void test1()
{
size_t i;
char cur;
unsigned char val;
for (i = 0; i < TESTDATALEN; i++) {
cur = testdata[i];
if (cur >= 97) {
val = cur - 97 + 10;
} else if (cur >= 65) {
val = cur - 65 + 10;
} else {
val = cur - 48;
}
/* even characters are the first half, odd characters the second half
* of the current output byte */
if (i%2 == 0) {
result[i/2] = val << 4;
} else {
result[i/2] |= val;
}
}
}
void test2()
{
size_t i;
char cur;
unsigned char val;
for (i = 0; i < TESTDATALEN; i++) {
cur = testdata[i];
val = base16_decoding_table1[(int)cur];
/* even characters are the first half, odd characters the second half
* of the current output byte */
if (i%2 == 0) {
result[i/2] = val << 4;
} else {
result[i/2] |= val;
}
}
}
void test3()
{
size_t i;
uint16_t *cur;
unsigned char val;
for (i = 0; i < TESTDATALEN; i+=2) {
cur = (uint16_t*)(testdata+i);
// apply bitmask to make sure that the first bit is zero
val = base16_decoding_table2[*cur & 0x7fff];
result[i/2] = val;
}
}
void test4()
{
size_t i;
uint16_t *cur;
unsigned char val;
for (i = 0; i < TESTDATALEN; i+=2) {
cur = (uint16_t*)(testdata+i);
val = base16_decoding_table3[*cur];
result[i/2] = val;
}
}
#define NUMTESTS 1000
int main() {
struct timespec before, after;
unsigned long long checksum;
int i;
double elapsed;
clock_gettime(CLOCK_MONOTONIC, &before);
for (i = 0; i < NUMTESTS; i++) {
test1();
}
clock_gettime(CLOCK_MONOTONIC, &after);
checksum = 0;
for (i = 0; i < TESTDATALEN/2; i++) {
checksum += result[i];
}
printf("checksum: %llu\n", checksum);
elapsed = difftime(after.tv_sec, before.tv_sec) + (after.tv_nsec - before.tv_nsec)/1.0e9;
printf("arithmetic solution took %f seconds\n", elapsed);
clock_gettime(CLOCK_MONOTONIC, &before);
for (i = 0; i < NUMTESTS; i++) {
test2();
}
clock_gettime(CLOCK_MONOTONIC, &after);
checksum = 0;
for (i = 0; i < TESTDATALEN/2; i++) {
checksum += result[i];
}
printf("checksum: %llu\n", checksum);
elapsed = difftime(after.tv_sec, before.tv_sec) + (after.tv_nsec - before.tv_nsec)/1.0e9;
printf("256 entries table took %f seconds\n", elapsed);
clock_gettime(CLOCK_MONOTONIC, &before);
for (i = 0; i < NUMTESTS; i++) {
test3();
}
clock_gettime(CLOCK_MONOTONIC, &after);
checksum = 0;
for (i = 0; i < TESTDATALEN/2; i++) {
checksum += result[i];
}
printf("checksum: %llu\n", checksum);
elapsed = difftime(after.tv_sec, before.tv_sec) + (after.tv_nsec - before.tv_nsec)/1.0e9;
printf("32768 entries table took %f seconds\n", elapsed);
clock_gettime(CLOCK_MONOTONIC, &before);
for (i = 0; i < NUMTESTS; i++) {
test4();
}
clock_gettime(CLOCK_MONOTONIC, &after);
checksum = 0;
for (i = 0; i < TESTDATALEN/2; i++) {
checksum += result[i];
}
printf("checksum: %llu\n", checksum);
elapsed = difftime(after.tv_sec, before.tv_sec) + (after.tv_nsec - before.tv_nsec)/1.0e9;
printf("65536 entries table took %f seconds\n", elapsed);
return 0;
}
$ gcc -O3 -g -Wall -Wextra test.c
并运行它:
$ ./a.out
void hex_binary(char * res){
char binary[16][5] = {"0000", "0001", "0010", "0011", "0100", "0101","0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110","1111"};
char digits [] = "0123456789abcdef";
const char input[] = "a9e6"; // input value
res[0] = '\0';
int p = 0;
int value =0;
while(input[p])
{
const char *v = strchr(digits, tolower(input[p]));
if(v[0]>96){
value=v[0]-87;
}
else{
value=v[0]-48;
}
if (v){
strcat(res, binary[value]);
}
p++;
}
printf("Res:%s\n", res);
}
void printBin(unsigned int num){
char str[sizeof(num)*8];
char *p = str;
for(*p='0'; num; num/=2) { *p++='0'+num%2; } //store remainders
for(--p; p>=str; putchar(*p--)) {;} //print remainders in reverse
putchar('\n');
}
switch (charval) {
case '0': binval = 0;
case '1': binval = 1;
case '2': binval = 2;
case '3': binval = 3;
....
case 'a': binval = 10;
case 'b': binval = 11;
case 'A': binval = 10;
case 'B': binval = 11;
....
case 'f': binval = 15;
case 'F': binval = 15;
default: binval = -1; // error case
}
0123456789abcdef
),然后通过在其上使用strchr()
来获取数字的值,要么我只需使用一个char [256]
数组,并将字符/数字作为索引。这样你既可以使用二进制表示作为值,也可以跳过额外的转换。 - Mario这是我用来将十六进制转换为二进制的函数,逐字节进行转换。
void HexToBin(char hex_number, char* bit_number) {
int max = 128;
for(int i = 7 ; i >-1 ; i--){
bit_number [i] = (hex_number & max ) ? 1 : 0;
max >>=1;
}
}
以及对函数的调用:
void main (void){
char hex_number = 0x6E; //0110 1110
char bit_number[8]={0,0,0,0,0,0,0,0};
HexToBin(hex_number,bit_number);
for(int i = 7 ; i >-1 ; i--)
printf("%d",bit_number[i]);
printf("\n");
system("pause");
}
这里是MSDOS的答案:
01101110
Press a key to continue . . .
非常简单!
#include <stdio.h>
int main()
{
long int binaryNumber,
hexadecimalNumber = 0,
j = 1,
remainder;
printf("Enter any number any binary number: ");
scanf("%ld", &binaryNumber);
while(binaryNumber != 0) {
remainder = binaryNumber % 10;
hexadecimalNumber = hexadecimalNumber + remainder * j;
j = j * 2;
binaryNumber = binaryNumber / 10;
}
printf("Equivalent hexadecimal value: %X", hexadecimalNumber);
return 0;
}