Division Assembly in MSP430

assemblymsp430

I have been trying to implement an assembly routine in MSP430 for division. I already got the code for the division from the Horner Division Algorithm. The problem is that I only got the integer part. How I can get the fractional part from a division? For example: 5/2 = 2.5. How do I get the 0.5 part?

Best Answer

You should get the remainder.

To turn that into a fraction you first need to decide between fixed and floating point; and if fixed, how many fractional bits.

Fixed point is easy : if you decide you want 8 fractional bits, just divide 2^8 * remainder / denominator, and use the size of that operation's remainder to determine rounding.

In your example, that would give

(256 * 1) / 2 = 128 as your fractional part, i.e. 128 / 256 = 0.5

Or for 3 fractional (decimal) digits, just compute 10^3 * remainder / denominator.

For floating point you would use a floating point library; it's far too complex to seriously consider rolling your own.

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Electronic – How critical a resistor value for MSP430 spy-bi-wire on reset

I would expect that several factors will put constrains on the size of the pull up resistor:

It must be big enough such JTAG driving circuit (programmer/debugger) will be able to override it,
It must be small enough that leakage current/noise will not change logic value of the line,
The RC constant that remove /RST after supply ramp up will have to be chosen to ensure proper reset.

Check documentation of uC and JTAG programmer for leakage and drive strength. Based on that size your pull up resistor. I would try to keep RC constant the same which would mean re-sizing capacitor. Note however that schematic you posted put maximum restriction on the capacitor to 2.2nF.

Standard disclaimer: You should follow manufacturer recommendation. There is good reason for that as good gals and guys that made the chip may have put really strange things into it (that are just waiting to bite you). What is worst such things may not be documented as describing them would reveal trade secret. If needed contact your support channel (usually not possible for small customers or DIYers - in this case experiment!). Note that I am not affiliated with TI and I do not have "inside" knowledge of the device.

Electronic – Assembly Language Program Design

Write a loop to shift the word left or right (doesn't matter) until the word is all zeros. The carry flag gives you the value of the next bit.

If carry is 1 and the previous carry was zero you're starting a new sequence of 1s. Then set a previous-was-a-first-one flag. (I assume that by contiguous you mean at least 2.)

If carry is 1 and previous-was-a-first-one is set you have a contiguous series, and increment your counter. Clear the previous-was-a-first-one flag.

If carry is 0 then clear the previous-was-a-first-one.

edit
Apparently "contiguous" doesn't require more than 1 bit, and then it's even simpler:

Set `previous' to `0`.
Shift left until word is all zeros.
If `carry` = `1` and `previous` = `0` increment counter.
Set `previous` to `carry`.

Best Answer

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