Author Topic: Parsing files  (Read 3788 times)

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Offline AngelFish

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Parsing files
« on: January 23, 2011, 03:56:01 pm »
How would one go about parsing a file in Axe? You'd clearly have to find the program's token in the VAT (Input?), but once you've found the pointer, how would you read the file itself? Would you simply treat it as a string and step through it or would you need to do something more difficult?
∂²Ψ    -(2m(V(x)-E)Ψ
---  = -------------
∂x²        ℏ²Ψ

Offline squidgetx

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Re: Parsing files
« Reply #1 on: January 23, 2011, 03:58:51 pm »
If by "file" you mean a program or appvar, then yes you can treat it as a string. Programs/appvars are pure data. What are you trying to do?

Offline AngelFish

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Re: Parsing files
« Reply #2 on: January 23, 2011, 04:16:40 pm »
Malbolge interpreter  ;D
∂²Ψ    -(2m(V(x)-E)Ψ
---  = -------------
∂x²        ℏ²Ψ

Offline Michael_Lee

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Re: Parsing files
« Reply #3 on: January 23, 2011, 04:22:45 pm »
Is anybody going to be able to actually write a program in Malbolge?
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Offline AngelFish

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Re: Parsing files
« Reply #4 on: January 23, 2011, 04:23:21 pm »
If you want working code, then Malbolge is not for you  :P
∂²Ψ    -(2m(V(x)-E)Ψ
---  = -------------
∂x²        ℏ²Ψ

Offline jnesselr

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Re: Parsing files
« Reply #5 on: January 23, 2011, 04:34:53 pm »
If you want working code, then Malbolge is not for you  :P
Yeah, Malbolge is one horribly painful programing language to write in.  Why just last week, I was trying to write an operating system in it. lol, imagine writing an operating system where the asm code was Malbolge.  That makes my head hurt just thinking about it.

Offline AngelFish

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Re: Parsing files
« Reply #6 on: January 23, 2011, 04:38:05 pm »
If I get to it, I might decide to make it even harder to program in  >:D
∂²Ψ    -(2m(V(x)-E)Ψ
---  = -------------
∂x²        ℏ²Ψ

Offline jnesselr

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Re: Parsing files
« Reply #7 on: January 23, 2011, 05:10:05 pm »
If I get to it, I might decide to make it even harder to program in  >:D
Is that even possible? Also, make sure that it has to be encrypted based on the hash of the programs variables 323 steps into the program.  If it doesn't do 323 steps, then they are modded. eg, if it only does 21 steps, then the 8th instruction is executed because 323 mod 21 = 8.  So The encryption is based around the number 8 somehow.

Offline AngelFish

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Re: Parsing files
« Reply #8 on: January 23, 2011, 05:11:03 pm »
Well, the encryption is rather simple and computers can obviously program in it. It can be made harder.
∂²Ψ    -(2m(V(x)-E)Ψ
---  = -------------
∂x²        ℏ²Ψ