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, 0x62, 0x6e,
0x66, 0x63, 0x51, 0x01, 0x0b, 0x55, 0x03, 0x6c,
0x69, 0x62, 0x6e, 0x66, 0x63, 0x2e, 0x6f, 0x72,
0x67
};
uint8_t *ndef_msg;
char *ndef_input = NULL;
How can I convert ndef_input
(which is just a plain text, like "hello") to hex and save into ndef_msg
?
As you can see ndef_default_msg
is in hex form. Data inside ndef_msg
should be something like that as well.
A bit of background, in the original program (source code), the program will open a file, get the data and put it inside ndef_msg
, which then will be written into a card. But I don't understand how it can take the data and convert to hex.
I want to simplify the program so it will directly ask user for text (instead of asking for a file).
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.