VirtualWire

Hey friend,

I am very confuse and lack of knowledge in using the virtualwire library. Can anyone guide me ?

I dont understand this : vw_send((uint8_t *)msg, strlen(msg));

What is that ? I want to ask, how if I want to just send a number to be received by the receiver ?

For example I want to send a number ‘2’. So in the receiver code, I can write like if the message received is ‘2’ then blink LED ?

Thanks :smiley:

I dont understand this : vw_send((uint8_t *)msg, strlen(msg));

"msg" is a null-terminated C string, probably in a "char" buffer.
So, "strlen" tell you how many characters are in the string (not including the terminator).
"(uint8_t *)" says to the compiler "look, I know I'm not giving you the exact type of pointer (msg) you expect, but it'll be OK. Trust me".
And vw_send sends the buffer.

If you want to send just '2' in this way, vw_send((uint8_t *)"2", 1);

Still not very clear..is that all about string , data type?

There is no string datatype in C - strings are just "char" arrays, with the last position filled with a zero (A true zero, not an ASCII zero).
So, the string "Hello" occupies six consecutive locations in array, 'H', 'e', 'l', 'l', 'o' with '\0', where the backslash signifies to the compiler that is really is zero, not ASCII zero.

So if I use this function vw_send(message, length), the messagee must be an arrary of 8 bits ? And I must include write it like this :
vw_send((uint8_t *)msg, strlen(msg)); ???

What are the alternatives ?

Thanks

the messagee must be an arrary of 8 bits ?

This doesn't make sense. The array can be any number of elements. The size of an element, in bytes, is dependent on the type of the array. All bytes are 8 bits.

If I want to just send number '2',

so can I just write vw_send(2, 1) ?

1 refer to the length of message .

so can I just write vw_send(2, 1) ?

No. For 2 reasons. First '2' is not equal to 2. Second, 2 is an int, not an array, and vw_send only sends arrays.

what do you mean 2 is not equal to 2 ? LOL

So everytime I write vw_send(I must include 'uint8_t', length) ?

Thanks

what do you mean 2 is not equal to 2 ? LOL

The number 2 and the letter ‘2’ are not equal.

You mean '2' and 'two' ?

You mean '2' and 'two' ?

No, I mean 2 as in 1+1 and the letter '2', between the letter '1' and the letter '3'.

The letter '2' has an ASCII code of 50. So, if you send '2', the value 50 gets sent. If you send 2, the value 2 gets sent.

Ok, I get you. It all about the usage of the ' ' .

So, everytime I sent a message using vw_send, I have to type it like this : vw_send((uint8_t*)msg, length ) ?

So, everytime I sent a message using vw_send, I have to type it like this : vw_send((uint8_t*)msg, length ) ?

Yes.

And I have to define what is msg ?

For example msg = '2' ?

PaulS:

So, everytime I sent a message using vw_send, I have to type it like this : vw_send((uint8_t*)msg, length ) ?

Yes.

Any alternatif ? And why got a * ?

Because you're casting a pointer.

Okay, I think I have to read more about pointer and array. Thanks !

vw_send(message, length)
Transmit a message. "message" is an array of the bytes to send, and "length" is the number of bytes stored in the array. This function returns immediately and the message is sent slowly by an interrupt-based background process.

so the message is refering to ((uint8_t*)msg), where msg is any number of words ? It is not array anymore because we include the uint8_t already ?

Thanks

For example msg = '2' ?

No.
2 != '2' != "2"

It is not array anymore because we include the uint8_t already ?

I don't understand what you're saying, but obviously it still is an array, because if it were not, then we wouldn't need to specify a buffer length to transmit.

I mean I can also write like this vw_send(char msg[8],8) ? Instead of writing it like vw_send((uint8_t*)msg,8), char msg[8] = {‘1’,‘2’,‘3’,‘4’,‘5’,‘6’,‘7’,‘8’}

Hahaha… well, so confuse ><