How can i convert a char array of number to byte array? Example:
char *digit="3224833640520308023"//long long array
convert to:
uint8_t buff[256]= {0x2c, 0xc0, 0xe9, 0x1c, 0x32, 0xf1, 0x55, 0x37, 0};
(2c c0 e9 1c 32 f1 55 37)
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If you put your code in a separate paragraph and indent each line by at least four spaces, it'll format much nicer and it'll be much easier for people to answer your question.Tim Martin– Tim Martin2011-01-26 16:48:49 +00:00Commented Jan 26, 2011 at 16:48
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What is the relationship between the contents of the char array and the contents of your uint8_t array?Oliver Charlesworth– Oliver Charlesworth2011-01-26 16:52:35 +00:00Commented Jan 26, 2011 at 16:52
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I take it you want to accept a string representing a number, convert to long long, and then have it as a uint8_t array with the binary representation? Or do you want to automatically convert it to a C statement like the one you've listed? Do you mind indulging in what is technically undefined behavior?David Thornley– David Thornley2011-01-26 16:53:33 +00:00Commented Jan 26, 2011 at 16:53
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Is your "digit" a group of individual ASCII characters? A giant decimal string? A group of hex digits? Please clarify...Brad– Brad2011-01-26 16:55:19 +00:00Commented Jan 26, 2011 at 16:55
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Big number (3224833640520308023) convert to hex (2cc0e91c32f15537)mGuest– mGuest2011-01-26 16:57:22 +00:00Commented Jan 26, 2011 at 16:57
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2 Answers
I printed in reverse order in the end. You may want to endian swap if you need the array in that endian order.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
char *digit="3224833640520308023";
int main() {
int i;
unsigned char byteArray[16];
unsigned long long x = strtoull(digit,0,10);
printf("%llx\n",x);
printf("%llu\n",x);
for (i=0;i<8;i++) {
byteArray[i] = (x>>(i*8)) & 0xFF;
}
printf ("Array is:\n");
for (i=7;i>=0;i--) {
printf("%2.2x ",byteArray[i]);
}
return 0;
}
3 Comments
Foo Bah
Instead of explicitly creating each value, you can cast a reference to an array. reinterpret_cast<uint8_t *>(&x)[i] (assuming endianness) should do the trick
mGuest
my device print to display: 2cc0e91c 750840092 Array is: 00 00 00 00 2c c0 e9 1c
Brad
Well, yes - this was C (not C++) so the "reinterpret_cast" won't work - but I could have "(unsigned char *) (&x)"
strtoull converts the string to a 64 bits internal representation.
htobe64 will switch the endianness to big endian (the one you used in your example) if needed on your platform.
You can then copy 8 bytes from this big endian 64bit variable to your byte array.
#include <stdint.h>
#include <endian.h>
#include <stdio.h>
#include <string.h>
char *digit="3224833640520308023";
main ()
{
uint64_t ull;
uint64_t beull;
uint8_t buff[8];
int i;
ull=strtoull(digit,0,10);
beull=htobe64(ull);
memcpy(buff,&beull,8);
for(i=0;i<8;i++)
{
printf("%02x\n",buff[i]);
}
}
