setting values and percentages

[EarlyBird2]

One thing I would like to know is what the */ function means

This is what I understand or misunderstand possibly.

Internally the maths can have a 32 bit result because the PIC has a 32 bit register. But if your maths calculation will fit in a word then using word=byte*byte will work and you do not need ** or */.

If you have wordH=word**word and wordL=word*word then you have all 32 bits stored in two words.

*/ is less obvious but gives the middle word of the 32 bits. Which is the same as word=word*word/256 losing the top and bottom 8 bits.

If as in your case and as you know 4095/100 gives a result of 40 therefore B_max = (4095/100)*maxbright will give 2000 when maxbright is 50.

Whereas 4095*256/100 gives result of 10483 and 50*/10483 gives 2047 because by multiplying by 256 and using the */ to remove the 8 lower bits (or divide by 256 which is the same) increased accuracy is achieved. So where is the precision lost? The division /100 is the source of lost precision so as a rule multiply first and divide last which is exactly what */ does in that your program multiplies by 256 and */ divides by 256 last. As an aside working this way reduces the calculations within the program, there is only a division

A question then. What is the result of 50*4095/100 2000 or 2047? and (4095/100)*50?

[richard]

result 50*4095/100 is 2047
result (4095/100)*50 is 2000

the calc follows typical bodmas rules (brackets, of , divide .....) see precedence in manual

the only 'suprising' bit is that the indermediate result (50*4095) is 204750 way too big for a word var. the pbp compiler code does indeed use 32 bit intermediate results.
the */ and ** operators allow you access these 32 bit results (providing you follow the rules of course)

[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