Re: [Pharo-users] why is masses not found?

Thu, 02 Jan 2020 05:42:22 -0800

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Hello Ben.

That was a error . There schould be a caret for it.  Im talking about the solution method.

No, I cannot clean the in part because it is used in the process*  methods

The  processData:  can also be deleted . I think Its  a part of a earlier try to solve it.

and yes. the ram is the same at both places. Is there then a better way to pass the stream around.

and of course thanks for the feedback.

Roelof



Op 2-1-2020 om 14:27 schreef Ben Coman:
Hi Roelof,

I presume its working since you only asked for code smells.   
It looks pretty good to me with methods generally being small and doing one thing.    

The way your class-side-method "IntComputer class >> solution" just creates an instance and sends #process is good.
I'm a little curious that it doesn't return a value.  #process returns a value but nothing is done with it.  I guess the test-frame doesn't require it.

The way zero-index ram access is encapsulated by  #at:  and   #at:put:  is good.

Possible cleanup,  #in and #in: don't seem required since variable "in" is used directly in the rest of the code, which is fine.

The hardcoded  #masses  return value is fine, except it makes the "masses" class variable redundant.

Your various #processXXX methods are succinct and consume the stream neatly.

However its curious your  #processData and #processData:  methods have identical code, since the latter has an argument that is then not used.
its just lucky that the "self ram" you pass there to #processData:  is the same as the "ram" variable referenced 

Well done.
cheers -ben





On Thu, 2 Jan 2020 at 23:37, Roelof Wobben via Pharo-users <pharo-users@lists.pharo.org> wrote:
Op 1-1-2020 om 17:27 schreef Roelof Wobben via Pharo-users:
Hello,

I made some changes. Is this better or are there still some code smells.
Code : https://github.com/rwobben/intComputer/blob/master/AOC%202019/IntComputer.class.st

if so, can someone help me to solve the code smells.

Here is the challenge I try to solve :

On the way to your gravity assist around the Moon, your ship computer beeps angrily about a "1202 program alarm". On the radio, an Elf is already explaining how to handle the situation: "Don't worry, that's perfectly norma--" The ship computer bursts into flames.

You notify the Elves that the computer's magic smoke seems to have escaped. "That computer ran Intcode programs like the gravity assist program it was working on; surely there are enough spare parts up there to build a new Intcode computer!"

An Intcode program is a list of integers separated by commas (like 1,0,0,3,99). To run one, start by looking at the first integer (called position 0). Here, you will find an opcode - either 1, 2, or 99. The opcode indicates what to do; for example, 99 means that the program is finished and should immediately halt. Encountering an unknown opcode means something went wrong.

Opcode 1 adds together numbers read from two positions and stores the result in a third position. The three integers immediately after the opcode tell you these three positions - the first two indicate the positions from which you should read the input values, and the third indicates the position at which the output should be stored.

For example, if your Intcode computer encounters 1,10,20,30, it should read the values at positions 10 and 20, add those values, and then overwrite the value at position 30 with their sum.

Opcode 2 works exactly like opcode 1, except it multiplies the two inputs instead of adding them. Again, the three integers after the opcode indicate where the inputs and outputs are, not their values.

Once you're done processing an opcode, move to the next one by stepping forward 4 positions.

For example, suppose you have the following program:

1,9,10,3,2,3,11,0,99,30,40,50

For the purposes of illustration, here is the same program split into multiple lines:

1,9,10,3,
2,3,11,0,
99,
30,40,50

The first four integers, 1,9,10,3, are at positions 0, 1, 2, and 3. Together, they represent the first opcode (1, addition), the positions of the two inputs (9 and 10), and the position of the output (3). To handle this opcode, you first need to get the values at the input positions: position 9 contains 30, and position 10 contains 40. Add these numbers together to get 70. Then, store this value at the output position; here, the output position (3) is at position 3, so it overwrites itself. Afterward, the program looks like this:

1,9,10,70,
2,3,11,0,
99,
30,40,50

Step forward 4 positions to reach the next opcode, 2. This opcode works just like the previous, but it multiplies instead of adding. The inputs are at positions 3 and 11; these positions contain 70 and 50 respectively. Multiplying these produces 3500; this is stored at position 0:

3500,9,10,70,
2,3,11,0,
99,
30,40,50

Stepping forward 4 more positions arrives at opcode 99, halting the program.

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