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Normal programming question

 
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stevenlkz



Joined: 05 Jun 2005
Posts: 5

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Normal programming question
PostPosted: Sat Jun 11, 2005 11:01 pm     Reply with quote

I got two questions about the programming of PIC16F877.
1. Normally, when we rotate 8 time with 8 bit data, we will get back the same data right? But if I rotate one time, then call a subroutine or goto other place, the carry flag already change. Then back to here and rotate again. This loop will go on for 8 time, is it finally still can get back the same data?
2. If I add 8 bit data with another 8 bit data and the result is is overload to 9 or 10 bit, then how a result will store?
PCM programmer



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Posts: 21708

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PostPosted: Sun Jun 12, 2005 12:09 am     Reply with quote

Quote:
Normally, when we rotate 8 time with 8 bit data, we will get back the same data right ? But if I rotate one time, then call a subroutine or
goto other place, the carry flag already change

If you use the CCS rotate functions instead of assembly language,
then you don't have to worry about this. If you make a test program,
compile it, and look at the .LST file, you will see the code created by
the compiler.

In the code below, you can see that CCS does the rotate operation
in two steps. First they rotate bit 7 of n into the Carry bit. They
don't care about the result, so they just stick it in the W register.
Their entire purpose is just to get bit 7 into the Carry bit.
Then in the next line, they do the actual rotation. They rotate n
to the left once, and the Carry bit is rotated into bit 0, and the
result is put back into n.
Code:

....................     rotate_left (&n, 1); 
0014:  RLF    n,W
0015:  RLF    n,F


In other words, the basic answer to many of your questions is:
Don't use ASM. Instead, use a built-in function. They have
already solved the whole problem for you. Do this:
Code:
 
int8 n;
rotate_left (&n, 1);


------------
Quote:
2. If I add 8 bit data with another 8 bit data and the result is
overload to 9 or 10 bit, then how a result will store?

You need to convert the 8-bit values to 16-bit before you do the addition.
This conversion is called "casting". Example:
Code:
int16 result;
int8 a, b;

result = (int16)a + (int16)b;
stevenlkz



Joined: 05 Jun 2005
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PostPosted: Mon Jun 13, 2005 12:18 am     Reply with quote

Thanks for help! But I really need to write in MPlab compiler. Sad.
I have another 6 question with MPlab assembler programming.
1. Is the PIC can use to check the value whether it is a positive or negative value?
2. Is the IF/ELSE statement available in MPlab?
3. How to compare and get the smallest number between three number?
4. After I use the fixed point division of the application note, AN617 in microchip, but the output is in floating point, then how to read the result?
5. If the maximum number of floating point is 0.125 in decimal number, then how many bits should I use for math operation?
6. In application note, AN575 in microchip, it said that floating point 0X823C5198, but how to read it? What decimal value is it?
PCM programmer



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PostPosted: Mon Jun 13, 2005 12:31 am     Reply with quote

I think you should study the C language.
Devin Electronics
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PostPosted: Fri Sep 16, 2005 1:25 pm     Reply with quote

I know C but lack knowledge on my Assembly so hofefully this is a simple question.

I have a 4 byte integer that I need to rotate out to the CARRY flag one bit at a time. I know I can use rotate_left(&msg_to_xmit, 4) easily enough but due to timing requirements, I only have one clock cycle at a time to do anything so I wrote the code to address that issue as shown in the example:

Code:
#inline
void transmit_0(void){
   PHASE_B_PIN = 1;                  //Turn on Phase B pin for two
   delay_cycles(1);                  //  cycles total.
   PHASE_B_PIN = 0;                  //Turn off Phase B pin.
   PHASE_A_PIN = 1;                  //Turn on Phase A pin for two
   delay_cycles(1);                  //  cycles total.
   PHASE_A_PIN = 0;                  //Turn off Phase A pin.
}

#inline
void transmit_0_shift_a(void) {
   PHASE_B_PIN = 1;                  //This is the same as above
   rotate_left(&msg_to_xmit+0,1);         //  except we use the NOPs
   PHASE_B_PIN = 0;                  //  to shift our next bit into
   PHASE_A_PIN = 1;                  //  position for our next
   rotate_left(&msg_to_xmit+1,1);         //  transmission.
   PHASE_A_PIN = 0;                  //
}

#inline
void transmit_0_shift_b(void) {
   PHASE_B_PIN = 1;                  //This is the same as above
   rotate_left(&msg_to_xmit+2,1);         //  except we use the NOPs
   PHASE_B_PIN = 0;                  //  to shift our next bit into
   PHASE_A_PIN = 1;                  //  position for our next
   rotate_left(&msg_to_xmit+3,1);         //  transmission.
   PHASE_A_PIN = 0;                  //
}


Now I find out that the rotate_left command take TWO cycles instead of one cycle that I would get from the assembly RLF command. If I re-write the code in assembly as such
Code:
#inline
void transmit_1_shift_a(void) {
   #asm
      BSF   PHASE_A_PIN                  //This is the same as above
      RLF msg_to_xmit+0, F            //  except we use the NOPs
      BCF PHASE_A_PIN                  //  to shift our next bit into
      BSF PHASE_B_PIN                  //  position for our next
      RLF msg_to_xmit+1, F            //  transmission.
      BCF PHASE_B_PIN                  //
   #endasm
}


it doesn't like my RLF msg_to_xmit+1,F line. Is using INDF/FSR the only option I have? And if so, is this the proper way to implimant it?
Code:
#inline
void transmit_1_shift_a(void) {
   #asm
      BSF PHASE_A_PIN                  //This is the same as above
      RLF FSR+0, F            //  except we use the NOPs
      BCF PHASE_A_PIN                  //  to shift our next bit into
      BSF PHASE_B_PIN                  //  position for our next
      RLF FSR+1, F            //  transmission.
      BCF PHASE_B_PIN                  //
   #endasm
}

void calling_routine(void) {
        #asm
                MOVLW msg_to_xmit
                MOVWF FSR
                CLRF INDF
        #endasm
        transmit_1_shift_a();
        transmit_1_shift_b();
}
Mark



Joined: 07 Sep 2003
Posts: 2838
Location: Atlanta, GA

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PostPosted: Fri Sep 16, 2005 1:58 pm     Reply with quote

This doesn't seem right
Code:
BSF   PHASE_A_PIN   

You aren't specifying the bit to set.
Guest








PostPosted: Fri Sep 16, 2005 2:19 pm     Reply with quote

PHASE_A_PIN is defined as follows:
Code:
struct port_layout{
   unsigned   p0:1;
   unsigned   p1:1;
   unsigned   p2:1;
   unsigned   p3:1;
   unsigned   p4:1;
   unsigned   p5:1;
   unsigned   p6:1;
   unsigned   p7:1;
};
struct port_layout port_a;
#byte port_a = 5

#define   PHASE_A_PIN      port_a.p0


And complies as:
Code:
BSF 0x5, 0
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