How to convert char to hex stored in uint8_t form?

Question

Suppose I have these variables,

const uint8_t ndef_default_msg[33] = {
    0xd1, 0x02, 0x1c, 0x53, 0x70, 0x91, 0x01, 0x09,
    0x54, 0x02, 0x65, 0x6e, 0x4c, 0x69         


        

Answer 1

Why not read it into ndef_msg directly, (minus the \0 if it suppose to be a pure array). The hex are just for presentation, you could have just picked decimal or octal with no consequence for the content.

void print_hex(uint8_t *s, size_t len) {
    for(int i = 0; i < len; i++) {
        printf("0x%02x, ", s[i]);
    }
    printf("\n");
}

int main()
{
    uint8_t ndef_msg[34] = {0};

    scanf("%33s", ndef_msg);
    print_hex(ndef_msg, strlen((char*)ndef_msg));


    return 0;
}

You probably need to handle the reading of the string differently to allow for whitespace and perhaps ignore \0, this is just to illustrate my point.