converting char* to array of uint8_t (c) -


i can make function calls , receive array of strings represents ipv6 adress. looks this

char* buffer=resolver_getstring(config, ini_boot_meshintfipaddress);

if printed buffer gate ipv6 adress in string form:

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

however, way how ipv6 address represented in project 16 hexadecimal number using uint8_t datatype follows

uint8_t ipadress[16]

now problem how can cast (or copy memory of buffer) uint8_t[16]

what 

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

is there anyway ? regards,

#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]);

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