Hex to Decimal

[proton]

How would i go about converting a large Hexidecimal number to Decimal in flowcode

IE. 1400DE15d1

Rgds

Wayne

[Steve]

The number you gave equates to a decimal value of 85913900497 and would need 5 bytes to store it (which is beyond Flowcode's current limits).

Another approach you could use it to use Binary Coded Decimal, in which case the conversion is relatively trivial.

[proton]

Is it possible to convert the HEX and only store the last 8 digits as i do not need the first 3 (859)

Rgds

Wayne

[Steve]

You would still need to perform the full conversion and then disregard the most significant 3 digits. If you explain what you are trying to do, we may be able to offer a better solution.

[proton]

I am trying use a RFID E-block to read are existing ID cards to log employees into a area and switch on the lights and only switch off the lights when the area is vacated

Using the reader it comes back with the following value from the 1st 5 bytes

20 0 222 21 209

Now to get the employees ID number i would have to convert this to HEX giving

1400DE15D1

then back to Decimal

85913900497

And then the last 8 Digits gives you the Employees number

13900497

Rgds

Wayne

(i have attached the flowcode)

Door_entry.fcf

(6.5 KiB) Downloaded 432 times

[Steve]

I see. You do need to look at the whole number to get that info. But I can suggest a shortcut plus a potential solution...

First the shortcut - you do not need to convert to hex. If you receive "20 0 222 21 209", then this equals 209 + (21*256) + (222*256*256) + (0*256*256*256) + (20*256*256*256*256) = 85913900497.

Second, a potential solution. Create a byte array of 14 bytes (I'll call this id[14]), then follow this procedure:

1) take the low byte (209) and break it into it's powers of 10 (i.e. 2, 0, and 9) and add these to the id locations id[11], id[12] and id[13]. At this stage, the id variable would contain: 0,0,0,0,0,2,0,9

2) take the next byte (21) and multiply it by 256 (to give 5376). Your holding variable will need to be an integer rather than a byte for this. Add these to the relevant id digits, which would result in "id" as follows: 0,0,0,0,0,0,0,0,0,0,5,5,7,15.

3) now cycle through from id[13] to id[0] and if the number is above 9, take 10 away and add 1 to the next number. This would leave "id" as 0,0,0,0,0,0,0,0,0,0,5,5,8,5.

4) unfortunately, the precision of numbers available within this version of Flowcode now runs out. So you need to use a trick for this and subsequent calculations (in fact, it would make sense to use this trick for previous calculations as well, even though you do not need to). As an example, I will take the final number which you would need to multiply by 4 lots of 256 (i.e. 20*256*256*256*256)...

a) take the number and multiply by 256 (which will give 5120) - break this into the powers of 10 like you have done before, this time into an INT array (say temp[14]).

b) multiply each of these digits by 256 each, giving ...,1280,256,512,0

c) then perform the decimal carries similar to step (3) above. For example: when working on the 512 entry, the "5" gets added to the 1280, the "1" to 256, and "2" stays in the same position. So, you would eventually end up with ...,1,3,1,0,7,2,0.

d) you then need to repeat steps (b) and (c) another 2 times. At this point, you will end up with the result of 20*256*256*256*256, i.e. ...8,5,8,9,9,3,4,5,9,2,0

e) finally, add each of these to the existing values in your "id" variable.

I know the procedure is not complete, but I hope you can see the methodology.

[proton]

Steve,

How would i go about intergrating the shortcut calculation in to my Flowcode program, program is post on above reply

Rgds

Wayne

[Steve]

Sorry, Wayne.

I've got loads of work on here at the moment. If I get any free time at home, I'll see if I can knock together some example macros that will make this procedure easy - but I can't promise anything.

[proton]

Thanks steve,

Every time i do the shortcut sum i get 85930677713 ?

What am i doing wrong

Rgd

Wayne

[Steve]

I've just tried it again and I get the original answer.

[proton]

Hi,

can anyone in this forum help me with this calculation as i,m stuck

tag4 + (tag3*256) + (tag2*256*256) + (tag1*256*256*256) + (tag*256*256*256*256) = 85913900497.

Tag to tag4 are variable data

example
tag = 20
tag1 = 0
tag2 = 222
tag3 = 21
tag4 = 209

Thanks

[proton]

Please can someone give me hand with the above, so i can put my E-blocks to some use

Pleading

Proton

[proton]

Is there any information on this site on how to creat a macro, so i can have a go at doing this calculation.

thanks

wayne

[Benj]

Hello Wayne

In Flowcode click on Help and Contents, Then scroll down to the section entitled working with macros and variables and look at the relevant help topics. Hopefully this should get you up to speed with using macros.

[proton]

Benj,

The answer to my calculation is larger than 1 byte, so what type of varible do i use. INT is ok for the first calculation, but when i get to tag2*256*256 it comes back as zero

Rgds

Wayne

[Steve]

Wayne,

I'm afraid what you are trying to do is not straight-forward. You will need to use a methodology similar to that in my previous post.

[proton]

Hi steve,

Sorry for pestering you, But how do i break down the 209 (tag) into powers of ten. I will get my head round this !

Rgds

Wayne

[Benj]

Hello

To break a variable into powers of ten you will have to do something similar to the following.

INT_Var = 10500

thousands = INT_Var MOD 1000 //should equal 10
INT_Var = INT_Var - (thousands * 1000)
hundreds = INT_Var MOD 100 //Should equal 5
INT_Var = INT_Var - (hundreds * 100)
tens = INT_Var MOD 10 //Should equal 0
INT_Var = INT_Var - (tens * 10)
units = INT_Var //Should equal 0