CODEWITHSUNDEEP
ccharipv6uint8t
lferasu
lferasu

Reputation: 185

converting char* to array of uint8_t (c)

I can make function calls and receive an array of strings that represents ipv6 adress. it looks something like this

char* buffer=resolver_getstring(config, INI_BOOT_MESHINTFIPADDRESS);

if i printed buffer i will gate the ipv6 adress in string form:

dddd:0000:0000:0000:0000:0000:0000:cccc

however, the way how ipv6 address is represented in my project is with 16 hexadecimal number by using uint8_t datatype as follows

uint8_t ipadress[16]

now my problem is how can i cast (or copy the memory of buffer) to uint8_t[16]

what i would like to get is 

    ipadress[0]=dd // hexadecimal number
    ipaddress[1]=dd
    ....
    ipaddress[15]=cc

is there anyway i could do ? Regards,

Upvotes: 2

Views: 2674

Answers (2)

BLUEPIXY
BLUEPIXY

Reputation: 40145

#include <stdint.h>
#include <inttypes.h>
...
char *buffer="dddd:0000:0000:0000:0000:0000:0000:cccc";
uint8_t ipadress[16];
sscanf(buffer,
    "%2" SCNx8 "%2" SCNx8 ":"
    "%2" SCNx8 "%2" SCNx8 ":"
    "%2" SCNx8 "%2" SCNx8 ":"
    "%2" SCNx8 "%2" SCNx8 ":"
    "%2" SCNx8 "%2" SCNx8 ":"
    "%2" SCNx8 "%2" SCNx8 ":"
    "%2" SCNx8 "%2" SCNx8 ":"
    "%2" SCNx8 "%2" SCNx8 ,
    &ipadress[0],&ipadress[1],
    &ipadress[2],&ipadress[3],
    &ipadress[4],&ipadress[5],
    &ipadress[6],&ipadress[7],
    &ipadress[8],&ipadress[9],
    &ipadress[10],&ipadress[11],
    &ipadress[12],&ipadress[13],
    &ipadress[14],&ipadress[15]);

Upvotes: 2

mfro
mfro

Reputation: 3335

You'll need to look into strtok() (to break the string into pieces at the colon char), strtol() (to convert the pieces from hexadecimal form into binary) and some bit shifting and shuffling (to break the resulting 16 bit numbers into two separate bytes).

Upvotes: 0

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