How to change types of variables, array of char to numbers and back to array of char?

[Malanar]

Hello,
Im using LabWindows to acquire data from the serial port.
My serial port is getting the following data:
A30<cr>
B540<cr>
C1023<cr>
D78<cr>
The letters are to indicate which sensor is sending the data and the number is the data itself. The numbers will vary in time going from nothing (  ) to 1023, it doesnt get zeros. Im using the carriage return to identify the end of a data, thats the <cr>.
The problem is:
labview put this data after the ComRdTerm in a array of characters, so if my variable is defined as "char escrita[20]" the value of escrita after reading the port would be "A30". I need to work in the value "30" as number, so how can i put that value in a variable of type integer? The value "A" can be lost. After working with the number i need to put it back as a array of char so i can write it in a textbox, im using the command Update Textbox to show the number on screen.
Another problem:
When i use the command ComRdTerm, the carriage return (ASCII value 13) goes in the variable or not?
If it goes then the value of escrita would be "A30<cr>"?

Thanks,
Sandro

[Malanar]

Just complementing the above.
How would be if i want the array of char to change to type floating point instead of a integer and then back to array of char?

[markus kossmann]

You might use the C standard functions sscanf()/sprintf() or the functions  Scan()/Fmt() from NIs Formatting and IO Library.

[Malanar]

Thanks for the reply, but i have no idea of how to use those commands. Can you write a little example using them?

[RobertoBozzolo]

OK, to read the number from the buffer you may use

sscanf (readBuffer + 1, "%d", &number);   // This fills an integer variable with your data, skipping initial character

sscanf (readBuffer + 1, "%f", &number);    // This read back in a double variable

To create a string with your data to show in the text box, simply use the following:

sprintf (buffer, "%d", number);    // Writes the first number

sprintf (buffer, "%s\n%d", buffer, number);    // Writes the 2nd (3rd, 4th...) number to the string

In the above statements "%d" can be substituted with "%f" for floating points numbers.

As an alternative, you may want to look of the examples in the online help, looking into the "Formatting and I/O library" chapter of the user guide: these will drive you in the use of Scan and Fmt native functions.

sscanf and sprintf are very big and complex commands that can be customized to fulfill your needs: their explanation goes beyond the scpe of this forum and I strongly reccommend that you study them on a c-language manual in order to use them at the best.

Message Edited by Roberto Bozzolo on 11-28-2005 09:44 PM

[jr_2005]

You might find it easier to use:

number = atoi (readBuffer + 1);

Noting that number is an integer, which you can copy to a float/double type variable later, if you prefer.

JR

[Malanar]

Thanks, that solved the problem. Actually i used the atof command but it works the same way, simple and easy.
Now i have to return the floating point value to an array of char, do you know what is the command for that?

[jr_2005]

As Roberto suggested, you will need the sprintf() function. Something like:

char string [32];
float number = 1234;

sprintf (string, "%f", number);  // "%g" is an alternative if there is never a fractional part

There is a wide variety of formatting options, such as: number of decimal places; forced sign inclusion; engineering/scientific notation etc. You may need to read up on all the options for the function in a good ANCI C text book.

JR

[Malanar]

It worked, thanks.
Now the last and simpler question, how do i round the number?
Example:
I have the number 24.932743539
I wanto to round it to 24.9
only 1 decimal.

And just to check, the sprintf didnt show any results in screen, is that right, if it is then good thats the way i need it.

[LuisG]

You can use the floor() function in the ANSI C library to round a number. If you just wanted to round it to nearest whole number, you would do this:

    newNumber = floor (oldNumber + 0.5);

But since you want to round it with one decimal point, you can do this instead:

    newNumber = floor (10 * oldNumber + 0.5) / 10;

Luis
NI