IF...ELSEIF Questions

kf4mat

Hi all,

I am currently working on the WAM text chapter 3 and have a quick question about the if...elseif statement. I looked in the help file and it does not say wether or not what I am doing is considered correct so here goes. How many ELSEIF statements can you have? If you modify the code as directed on page 86 and push both buttons both LED's flash, however unless I am reading incorrectly it only contains code to flash the led on P14 if only one button is pushed (IN3).

My solution was to add another elseif statement, which worked, but is this how it should be done or is there a more elegant way to do this????

Tom

Here is a copy of my activity #4 code

' {$STAMP BS2}
' {$PBASIC 2.5}
DO
· DEBUG HOME
· DEBUG ? IN3
· DEBUG ? IN4
· IF (IN3 = 1) AND (IN4 = 1) THEN
· HIGH 14
· HIGH 15
· PAUSE 50
· ELSEIF (IN3 = 1) THEN
· HIGH 14
· PAUSE 50
· ELSEIF (IN4 = 1) THEN
· HIGH 15
· PAUSE 50
· ELSE
··· PAUSE 100
· ENDIF
· LOW 14
· LOW 15
· PAUSE 50
LOOP


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Jon Williams

You could use SELECT-CASE, but <EDIT> another variable to capture the inputs is suggested. </EDIT>· Since you're using IN3 and IN4, you could do it like this

· testIn = (INL & %00011000) >> 3
· SELECT testIn
··· CASE %11
····· HIGH 14
····· HIGH 15
····· PAUSE 50
··· CASE %01
····· HIGH 14
····· PAUSE 50
··· CASE %10
····· HIGH 15
····· PAUSE 50
··· CASE ELSE
····· PAUSE 100
· ENDSELECT
· LOW 14
· LOW 15
· PAUSE 50

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Jon Williams
Applications Engineer, Parallax

Post Edited (Jon Williams (Parallax)) : 10/16/2005 3:59:19 AM GMT

Fe2o3Fish

<mode picking="nits">
Have I missed something? Why is the extra variable is "required" in the SELECT statement?

SELECT ((INL & %00011000) >> 3)

is perfectly acceptable to the PBASIC 2.5 compiler. True, it's nice if you can spare the RAM
but when things get tight.... definitely not required.
</mode>

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-Rusty-
--
Rusty Haddock = KD4WLZ = rusty@fe2o3.lonestar.org
**Out yonder in the Van Alstyne (TX) Metropolitan Area**
Microsoft is to software what McDonalds is to gourmet cooking

Jon Williams

You're right ... you're picking nits! ·· All kidding aside, it is true that a variable is not required -- I suppose it's just a matter of style and with new programmers I tend to get VERY verbose.

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Jon Williams
Applications Engineer, Parallax

kf4mat

Well that's a bit above my head but, based on what I am expecting to happen my guess is that you tell the stamp to look for a 1 on P3 & P4 with the statement (INL & %00011000) >> 3. With the LSB in the word representing P0 and the MSB representing P7. Not quite sure what the >>3 does here or the INL, but I'm just learning so I won't panic over it. As far as the rest goes each case represents the status of the pushbuttons. It would appear that if I had three pushbuttons I could write the code as case %111 or case %101 ect. I print this out to refer back to once I learn some more, but for now since mine works correctly I'll leave it as is.

Thanks for the input,

Tom

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Jon Williams

>> is the shift-right operator.· Let's say IN3 and IN4 are both 1, other pins 0:

· temp = INL

At this point temp will be equal to %00011000.· Now if we do this:

· temp = temp >> 3

... temp will end up with %00000011.

See how the bits were shifted to the right?· There is a complimentary operator, <<, that shifts bits left.· On another note, shifting right by one bit is the same as dividing by two; shifting-left by one bit is the same as multiplying by two.

So...

· temp = temp >> 3

... is actually the same as:

· temp = temp / 8

... but shifting is more efficient (internally) than division.· Advanced programmers with look for power-of-2 division and multiplication opportunies to use shift operators.· You could replace:

· myValue = testInput * 32

... with:

· myValue = testInput << 5

... and the code will run a little faster.

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Jon Williams
Applications Engineer, Parallax

Post Edited (Jon Williams (Parallax)) : 10/16/2005 7:18:05 PM GMT

Bruce Bates

Tom -

Just add the following information to what Jon has already supplied, and you'll be able to see exactly what he did with his tricky bit of coding:
(INL & %00011000) >> 3

The pin ports on the PBASIC Stamp can be addressed individually (In0, In1, In2, etc for input), in four groups of four successive pins (INA, INB, INC, IND for input) or in two groups of eight successive pins (INL, INH or Input Low, Input High). Thus INL addresses Input pin ports 0-7 collectively, as one unit. What was said above about the INPUT pins is also true of the OUTPUT pin ports as well (Out1, Out2, ...; OUTA, OUTB, ...; OUTL, OUTH, etc).

If you want to set a bit ON, you use the bitwise OR operator (|) with a 1 in the mask position where you want the bit turned ON. If you want to set a bit OFF, you use the bitwise AND operator (&) and use a 0 in the mask position where you want the bit turned OFF. This can be used for single bits (a one bit mask) or multiple bits (a multi-bit mask). In this piece of coding, a multi-bit mask is being used ("%00011000"), 8 bits wide.

Thus, in the section of that instruction "INL & %00011000", the first operation (INL) is to field the status of Input pin ports 0-7 and place the result in an unseen, temporary variable. The second succesive operation "&" (bitwise AND) uses that same temporary variable as the source field, and sets everything EXCEPT the bits that represent the state of Input pin ports 3 and 4 to zero or OFF, since they are presently of no interest to us. The bits representing the state of Input pin ports 3 and 4·are left alone. In the portion of this instruction string we're looking at now, INL is the original source field, an unseen temporary variable is the receiving field for the first operation, and "%00011000" is the "bit mask" of which I'm speaking. The mask bits are applied to the temporary variable; thus it is the source field for the second operation.

As Jon said in his reply, in the last segment,or third operation,·of this instruction ">> 3" the result of all that, which is still contained within the temporary variable, is effectively "moved" to the right (shifted right) three bit positions, leaving the result in the 2 LSB positions of the unseen temporary variable, and zero filling the "emptied" MSB bits.

I hope that's helpful and not more confusing.

Regards,

Bruce Bates

Post Edited (Bruce Bates) : 10/16/2005 8:47:40 PM GMT

Andy Lindsay (Parallax)

Tom,

Your solution is correct.· It was the conclusion I was hoping you would draw based on what you have learned so far in What's a Microcontroller.

One improvement you can make to the·program you posted is simply to indent the code blocks·after each IF, ELSEIF, ELSE condition.· Here is an example:

' {$STAMP BS2} 
' {$PBASIC 2.5} 
DO 

 
  DEBUG HOME 
  DEBUG ? IN3 
  DEBUG ? IN4 

 
  IF (IN3 = 1) AND (IN4 = 1) THEN 

    HIGH 14 
    HIGH 15 
    PAUSE 50 
  ELSEIF (IN3 = 1) THEN 
    HIGH 14 
    PAUSE 50 
  ELSEIF (IN4 = 1) THEN 
    HIGH 15 
    PAUSE 50 
  ELSE 
    PAUSE 100 
  ENDIF 

 
  LOW 14 
  LOW 15 

 
  PAUSE 50 

 
LOOP

Notice that everything between DO and LOOP is indented by at least two spaces.· It helps the reader see that everything is·happening in a single DO...LOOP.· In addition, the code blocks that follow each (IF...ELSEIF...ELSE...) are indented two more spaces.· This makes it easier to see what action the program takes for each condition.

If you have more questions as you go through the activities in What's a Microcontroller, consider posting them to the Stamps in Class forum.· In addition to other members who have worked through or are working through the same books, most if not all of the·Stamps in Class authors monitor that forum for questions that pertain to their books.·

Regards, Andy

Post Edited (Andy Lindsay (Parallax)) : 10/17/2005 3:57:57 AM GMT

Andy Lindsay (Parallax)

Oh, one other suggestion. You can also replace the three DEBUG commands with one - DEBUG HOME, ? IN3, ? IN4

kf4mat

Thanks for all the help guys. I really appreciate it.

Tom



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