setting values and percentages

[richard]

however
if you have a word var say numx
and x =4095
then 50*numx/100 the result is 81
the word var causes a intermediate stage overflow (bits 0-15 of 204750 =8142)
I still say integer math is tricky and results need to be double checked

[EarlyBird2]

however
if you have a word var say numx
and x =4095
then 50*numx/100 the result is 81
the word var causes a intermediate stage overflow (bits 0-15 of 204750 =8142)

I do not understand this.
What is the relationship between numx and x?
Why does an intermediate stage overflow occur and what is an intermediate stage overflow?

[EarlyBird2]

I do not understand this.

Why does an intermediate stage overflow occur and what is an intermediate stage overflow?

Could you explain please?

[richard]

What is the relationship between numx and x?

poor typing
if you have a word var say numx
and numx =4095
then 50*numx/100 the result is 81
the word var causes a intermediate stage overflow (bits 0-15 of 204750 =8142)
I still say integer math is tricky and results need to be double checked

intermediate overflow
50*4095 =204750 max value for word 65535 therefore an overflow occurs and the result becomes 8142

[EarlyBird2]

intermediate overflow
50*4095 =204750 max value for word 65535 therefore an overflow occurs and the result becomes 8142

But the result is not stored in numx it is stored in another word variable called result?

[richard]

simply stated
if there is a var in a multilstage calculation then any intermediate results must be able to be stored in that var , its just the way it works.
try it yourself
the */ ,** ops are the only workaround

[richard]

you might have missed the point

50*4095/100 is 2047 this works (all constants)

numx=4095
50*numx/100 is 81 this does not

numy=50
numy*4095/100 is 81 this does not

numz=100
50*4095/numz is 81 this does not

[EarlyBird2]

Now I understand what you are saying. It does seem odd to me that there is a 32 bit register for maths calculation which can not be used if there is a word or byte variable in the calculation because the word/byte variable could overflow. A bit more thinking is required on my part, I think.

Does this mean that using ** and */ uses 32 bit and * only 16 bit?

[Dave]

Look at the Statement keyword "DIV32". It is possible to compute all of these statements:
numx=4095
50*numx/100

numy=50
numy*4095/100

numz=100
50*4095/numz

[EarlyBird2]

Now I understand what you are saying. It does seem odd to me that there is a 32 bit register for maths calculation which can not be used if there is a word or byte variable in the calculation because the word/byte variable could overflow. A bit more thinking is required on my part, I think.

Does this mean that using ** and */ uses 32 bit and * only 16 bit?

After more thought and a bit of direction to look at DIV32 by Dave.

Multiplication stores a 32 bit result internally but the normal divide only works on the lower 16 bits which causes the 'overflow' problem highlighted by Richard. DIV32 was created specifically to overcome the 16 bit division limitation and as Dave says is a solution here.

Scampy asked about */ and how that works. It works by removing the division in the calculation so there is no 16 bit divide, there is an 8 bit shift or a divide by 256 of the 32 bits whichever way one wishes to visualise the process. Knowing that */ will apply a division by 256 one has to take account of this in the program.

In this case

B_max = (4095/100)*maxbright

take 4095/100 and multiply by 256 giving 10483.2 which is rounded up to 10484. Resulting in this

B_max = maxbright*/10484 which results in an integer value of 2047 when maxbright is 50.