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sending a string of HEX via serial port.
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freefuel

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Joined: 27 Aug 2007
Posts: 55
Location: New Jersey, USA

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PostPosted: Fri Dec 21, 2007 4:08 pm    Post subject: sending a string of HEX via serial port. Reply with quote

I'm trying to send the HEX values of 02 00 00 00 07 out the serial port but I keep getting an error stating

Error : 242 Line : 18 Source variable does not match target variable [TEMP1 = 1 , 1 , 1 , 7]

why am I getting this error?

$crystal = 8000000
$baud = 38400
'tell the compiler which chip we use
$regfile = "m16def.dat"

'some variables we will use
Dim S As String * 5

S = String(2 , 1 , 1 , 1 , 7)

Dim Mybaud As Long

'when you pass the baud rate with a variable, make sure you dimesion it as a LONG

'Ucsr0b = 0 ' DISABLE HW UART

Mybaud = 38400

Do
Print "Hello World"
'first get some data
Serin S , 0 , D , 0 , Mybaud , 0 , 8 , 2
'now send it
Serout S , 0 , D , 1 , Mybaud , 0 , 8 , 2
' ^ 1 stop bit
' ^---- 8 data bits
' ^------ even parity (0=N, 1 = E, 2=O)
' ^-------------- baud rate
' ^-------------------- pin number
' ^----------------------- port so PORTA.0 and PORTA.1 are used
' ^--------------------------- for strings pass 0
' ^-------------------------------- variable
Wait 1
Loop
End

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pratik

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PostPosted: Fri Dec 21, 2007 7:46 pm    Post subject: Reply with quote

hi,
read the help for "String" command
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kimmi

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PostPosted: Fri Dec 21, 2007 9:23 pm    Post subject: Reply with quote

Hi freefuel ,
What uart do you use software uart ?
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Frankeman

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PostPosted: Fri Dec 21, 2007 9:37 pm    Post subject: Reply with quote

Hi,

Code:
S = String(2 , 1 , 1 , 1 , 7)
 

This is no valid sourcecode, fill an array with the values 2,1,1,1,7 and send that away.

Frank.
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freefuel

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PostPosted: Sat Dec 22, 2007 10:38 am    Post subject: Reply with quote

this is a sample directly from the help file.

when I compile that it passes.

Dim S As String * 15

S = String(5 , 65)

Print S 'AAAAA

End

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Frankeman

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PostPosted: Sat Dec 22, 2007 10:52 am    Post subject: Reply with quote

Try to compile with the line:
S = String(2 , 1 , 1 , 1 , 7)

And then read the help about the string command.
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freefuel

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PostPosted: Sat Dec 22, 2007 11:20 am    Post subject: Reply with quote

I did and I get the error listed in my first post.
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Evert :-)

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PostPosted: Sat Dec 22, 2007 12:41 pm    Post subject: Reply with quote

Syntax
var = STRING(m ,n)

Var
The string that is assigned.

N
The ASCII-code that is assigned to the string.

M
The number of characters to assign.

Code:

Dim S As String * 15
S = String(5 , 65)
Print S                                                     'AAAAA
 


This sample will print 5x the ascii karakter 65 = A, as you can see here in the sample it wil print AAAAA.

What you want to do with STRING is not possible, you need something like this.
S = Chr(&H02) + Chr(&H00) + Chr(&H00) + Chr(&H00) + Chr(&H07)

OR

S = "{002}{000}{000}{000}{007}"

And what the rest it trying to say to you that you must read the helpfile, there's very clear in the help file that STRING only allows 2 parameters.


MERRY CHRISTMAS TO YOU ALL Laughing

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freefuel

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PostPosted: Sat Dec 22, 2007 12:58 pm    Post subject: Reply with quote

Thank you Evert

after seeing your example I now see that string command can only send the same charter.

I tried looking for an array command but could not find one.

Is there a way to store are retrieve a series of different values?

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pratik

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PostPosted: Sat Dec 22, 2007 5:00 pm    Post subject: Reply with quote

Hi,

try this


$crystal = 8000000
$baud = 38400
$regfile = "m16def.dat"

Const Cmaxchar = 160
Dim Tx(cmaxchar) As Byte
Dim I As Byte
Dim B As Byte

For I = 1 To 5

Read B
Tx(i) = B

Next I

Do

Print "Hello World"

For I = 1 To 5
Print Hex(tx(i))
Next I

Wait 1

Loop


Dta:
Data 02 00 00 00 07


by & MERRY CHRISTMAS TO YOU ALL
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Pixelfrank

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PostPosted: Mon Mar 31, 2008 4:28 pm    Post subject: Reply with quote

Hello,
i have an problem too, when i try to get five Nulls (&H00) into a String.

1. Disp1 = "{000}{000}{000}{000}{000}{001}FF00{002}

2. Disp1 = String(5 , &H00) + Chr(&H01) + "FF00" + Chr(&H02)

3. Disp1 = Chr(&H00) + Chr(&H00) + Chr(&H00) + Chr(&H00) + Chr(&H00) + Chr(&H01) + "FF00" + Chr(&H02)

in example 1 there is nothing in the string (???)
both example 2 and 3 contains the SOH (&H01) + "FF00" + STX(&H02), but the Nulls are ignored.
i want use the string to initilize an external RS232 device.
when i use:

4. Print Chr(&H00) ; Chr(&H00) ; Chr(&H00) ; Chr(&H00) ; Chr(&H00) ; Chr(&H01) ; "FF00" ; Chr(&H02) ;

this works fine, but it is not so "readable"

Frank
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DToolan

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PostPosted: Mon Mar 31, 2008 4:46 pm    Post subject: Reply with quote

Frank... a null character designates the end of a string so a normal print command will stop at the null character.

People are so hung up on using strings... *eyeroll*
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Pixelfrank

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PostPosted: Mon Mar 31, 2008 4:53 pm    Post subject: Reply with quote

Pixelfrank wrote:
Print Chr(&H00) ; Chr(&H00) ; Chr(&H00) ; Chr(&H00) ; Chr(&H00) ; Chr(&H01) ; "FF00" ; Chr(&H02) ;


Hello DToolan,
thanks for answering.

But why works this above?

For the external RS232 device it is inportant to get the 5 Nulls before.

Frank
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Beke

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PostPosted: Mon Mar 31, 2008 6:53 pm    Post subject: Reply with quote

Hi,
AFAIK, Chr(0) does not exist, it is the "null" character, i.e. nothing. To see the print commands behaviour with it, try the following code:
Code:
Dim A As String * 5
  A = Chr(65) + Chr(66) + Chr(67) + Chr(0) + Chr(68)
  Print A
  Print Len(a)
  End  

Strings are terminated by the 0 "value".
According to the Help, the binary content of a variable is printed by the Printbin command.(its pair is the Inputbin).
Why Pixelfrank's command works, I cannot explain. An oscilloscope could clarify the situation.
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DToolan

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PostPosted: Mon Mar 31, 2008 6:58 pm    Post subject: Reply with quote

Quote:
But why works this above?

From the Bascom Help...

CHR
Action

"Convert a numeric variable or a constant to a string with a length of 1 character."

"When you use PRINT Chr(numvar), the ASCII character itself will be printed."

This implies that when using CHR(&H00) that the print command will print it whether it is a null or not because it knows to stop after just one character.

When printing a normal string... ie, "HELLO WORLD" there is a null character at the end of "HELLO WORLD" and that is how it knows when to stop.

Since you are using CHR(&H00) it knows to print just one character and then stop.

You could just use Printbin and an array? Oh... thats not a string. People and strings... gotta' love them.

PixelFrank... you could also do the following
Code:
DIM wTemp As Word

wTemp = LOADLABEL(Init)

PRINTBIN wTemp

Do : Loop

End

Init:

DATA &H00 , &H00 , &H00 , &H00 , &H00 , &H01 , &H46 , &H46, &H30 , &H30 , &H00 , &H02


Last edited by DToolan on Mon Mar 31, 2008 9:08 pm; edited 2 times in total
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