Having trouble converting a binary string to decimals .... leading zeroes in a string 'disappear' and break the conversion?

Author

Message

PHeMoX

6

Years of Service

Recently Online

Joined: 9th Jan 2018

Location:

Posted: 25th Oct 2018 14:26 Edited at: 25th Oct 2018 14:27

Link

As the title says, I need some help with converting a binary string to decimals. My code works, but only when the binary string starts with a 1. If it starts with a 0, it gets confused lol.
I suspect this is due to how a string doesn't properly save 'leading zeroes'.

Any tips on how to fix this?

Bad;

Good;

Attachments

Login to view attachments

Back to top

Profile PM

Bengismo

7

Years of Service

User Offline

Joined: 20th Nov 2017

Location: Yorkshire, England

Posted: 25th Oct 2018 14:50

Link

lol... we would need to see some code otherwise... its just guessing

Back to top

Profile PM

PHeMoX

6

Years of Service

Recently Online

Joined: 9th Jan 2018

Location:

Posted: 25th Oct 2018 14:51 Edited at: 25th Oct 2018 14:58

Link

I think I have fixed it in terms of the math, by limiting the base it goes through, however the 'leading' zero in the final string that's printed is missing :

From what I can tell, the issue is in this piece:

+ Code Snippet

Print(" bin number = "+str(binary_val))
	Print(" dec number = "+str(decimal_val))
	[b]convert$ = bin(decimal_val)
[/b]    Print(convert$+" binary")

valu is just an array converted to string to integer; might also be the issue, but the math seems correct, so I'm confused:

+ Code Snippet

valu = val(str(orig_gene[0,0])+str(orig_gene[0,1])+str(orig_gene[0,2])+str(orig_gene[0,3])+str(orig_gene[0,4]))
			conv_bin(valu)

Here's the extremely basic conversion code:

+ Code Snippet

function conv_bin(num)
	while base <32  // 2x2x2x2x2= 32 ; string length 
			remm = mod(num,10)
			decimal_val = decimal_val + remm *base 
			num = num / 10
			base = base * 2
	endwhile
endfunction

Attachments

Login to view attachments

Back to top

Profile PM

PHeMoX

6

Years of Service

Recently Online

Joined: 9th Jan 2018

Location:

Posted: 25th Oct 2018 15:03

Link

Back to top

Profile PM

PHeMoX

6

Years of Service

Recently Online

Joined: 9th Jan 2018

Location:

Posted: 25th Oct 2018 15:09

Link

I guess Bin(number) doesn't output leading zeroes or are they simply ignored by Print() ?

Back to top

Profile PM

Bengismo

7

Years of Service

User Offline

Joined: 20th Nov 2017

Location: Yorkshire, England

Posted: 25th Oct 2018 15:17

Link

This post has been marked by the post author as the answer.

Bin ignores leading zeros.....otherwise you would have 32 digits every single time you used it

Most operations dont print leading zeros....int decimal values are not 0000023 etc....

If you do want a minimum number of digits from bin() then simply use something like

function binl(i as integer, digits as integer)

t$ = bin(i)
while len(t$)<digits
t$ = "0" + t$
endwhile

endfunction t$

That will give you a defined number of digits everytime

Back to top

Profile PM

PHeMoX

6

Years of Service

Recently Online

Joined: 9th Jan 2018

Location:

Posted: 25th Oct 2018 15:29

Link

Thanks! Yeah, I figured as much that Bin ignores leading zeros. Probably makes sense it does. But your code example fixes the issue, thank you.

Back to top

Profile PM

Carharttguy

8

Years of Service

User Offline

Joined: 8th Jul 2016

Location: Belgium

Posted: 25th Oct 2018 18:13

Link

Bin() just ignores de leading zeros as there are not significant, there is no way for Bin() to guess how muchs zeros you want.
You seem to expect 01000, some people want 00001000, some want 000000000000000000000000000000000000000000001000.
Just add them yourself if you need them is the best solution I think.

Back to top

Profile PM