C - converting char* to unsigned char array

Question

I'm trying hard to convert char *macAddress into unsigned char mac[6] but without success so far. My input string looks like "00:10:6f:16:01:b3" and I want to devide it into mac[0] = 0x00, mac[1] = 0x10, etc..

I've tried to use memcpy like:

memcpy(&mac, (unsigned char)macAddress, sizeof mac);

or same another ways like mac = (unsigned char *)macAddress and so on but nothing worked for me well.
Is there any correct way how to convert it without any precision loss?

(unsigned char)macAddress?? What makes you think you need or should cast a pointer to an unsigned char?? — UnholySheep
– UnholySheep, Commented Apr 20, 2017 at 21:19
memcpy is a C function. Where is you rminimal reproducible example and which language do you use? Remove the tag for the unrelated language. — too honest for this site
– too honest for this site, Commented Apr 20, 2017 at 21:22
What is in macAddress? Is it a string with colons like "00:10:6f:16:01:b3"? — stark
– stark, Commented Apr 20, 2017 at 21:28
Tell us what you're trying to do, not just what types the input and output have. — Lightness Races in Orbit
– Lightness Races in Orbit, Commented Apr 20, 2017 at 21:40
Four steps: 1) describe the structure of the data in your input; 2) describe the structure of the desired result; 3) describe the rule for transforming the data in your input into the desired result; 4) write the code. — Pete Becker
– Pete Becker, Commented Apr 20, 2017 at 21:40

Micha Wiedenmann · Accepted Answer · 2017-04-21 06:44:03Z

2

You want to convert hexadecimal digits (i.e. the string "00") of the mac address to byte values (i.e. the value 0). The memcpy instead copies the value of the string digits (i.e. '0' is 48 or 0x30).

You can use sscanf for the correct conversion.

At least on Linux, you can also use the function ether_aton. See the man page.

edited Apr 21, 2017 at 6:44

Micha Wiedenmann

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answered Apr 20, 2017 at 21:24

Ludwig Schulze

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