Файл: The quintessential PIC microcontroller (S. Katzen, 2000).pdf

6. Subroutines and Modules 157

Program 6.6 (continued.) Implementing a byte multiply using a stack model.

its use of scarce data memory in large software systems. However, in programs running on low and mid-range PICs are often not very complex. Furthermore the small Program memory (1024 instructions for the PIC16F84) may further restrict the use of this relatively extravagant technique. Where real-time execution time is critical the additional burden of stack handling is unlikely to be worthwhile.

Examples

Example 6.1

The binary series approximation to the fraction 13 is:

1 1 1 1 1 1 1 1 3 = 2 − 4 + 8 − 16 + 32 − 64 + 128 · · ·

Using this series, write a subroutine that will divide a byte in the Working register by three with the quotient being returned in the same W register. The actual value of the summation up to 1281 is 0.3359375, which is within 0.78% of the exact value. With an 8-bit datum there is no point in including any further elements in the series, but where 16-bit operands are being processed then further elements up to the desired accuracy can be summed in the same manner.

158 The Quintessential PIC Microcontroller

Solution

The fractions 12 , 14 , 18 etc. can be easily generated by shifting right. The coding listed in Program 6.7 simply repetitively shifts the number as copied to a temporary location in register file File 33h right. Notice that it is necessary to clear the Carry flag before each shift as both the previous Rotate and Subtract instructions can alter its state. As the subwf instruction subtracts the contents of W (the sub quotient) from the datum in the specified file register then the outcome being built up in W will oscillate in sign as desired. In situations where the series element signs

Program 6.7 Dividing by three

; ************************************************************

;************************************************************

;Local declarations

160 The Quintessential PIC Microcontroller

Solution

For a delay as long as this we need to extend Program 6.1 to use a larger count. In the coding of Program 6.9 three file registers are used to give a triple loop.

File register COUNT2 is initialized to the value H whilst the other two file registers are cleared giving an e ective count range of 256. The outer loop exits the program when COUNT2 reaches zero. This single pass therefore contributes (H × 3) − 1 cycles to the total delay. Each pass through the COUNT1 loop takes (256 ×3) −1 cycles and there are H passes giving a contribution of H × [(256 × 3) − 1] cycles. In the same manner the inner loop based on decrementing COUNT0 runs H × 256 times each pass contributing (256 × 3) − 1 cycles to the total. Thus we have a total delay of:

(H × 3) − 1 + H × [(256 × 3) − 1] + H × 256 × [(256 × 3) − 1] + 6

Equating this to 106 ( µs per second) gives H ≈ 5. The actual delay with an H of 5 is 0.985615 s which is accurate to 1.4%. If desired the shortfall of 14,385 cycles can be made up by adding nop instructions to the middle count loop. Each such instruction gives an extra 1280 cycles. The

Program 6.9 A 1-second delay program.

; *************************************************************

; *************************************************************

DELAY_1_S movlw movwf clrf clrf

D_LOOP

decfsz goto decfsz goto decfsz goto return

6. Subroutines and Modules 161

maximum delay possible with this program is 50.46 s for H = 0 (e ectively 256).

Example 6.4

Design a subroutine to convert a binary byte passed in W to a 3-digit BCD equivalent in HUNDREDS (File 30h), TENS (File 31h) and UNITS (File 32h).

Solution

We have already coded a routine to implement this mapping in Example 5.3 on page 129. However this was restricted to a range 0–99, that is two digits. Nevertheless we can extend the technique used there by first subtracting and counting hundreds from the original binary byte. After this has been computed then the residue will be less than 100 and the rest of the coding will be the same, as shown in Program 6.10. Thus a suitable task list would be:

1.Divide by 100; the remainder is the hundreds digit.

2.Divide the quotient by ten; the remainder is the tens digit.

3.The quotient is the units digit.

Program 6.10 Binary to 3-digit BCD conversion.

;************************************************************

;* FUNCTION: Converts a binary byte in W to three BCD digits*

;************************************************************

;First divide by a hundred