Re: [Pharo-users] can I write this without the three if then;s

The “Kent Beck approved solutions” here are two: 1) To use a dictionary (or ordered dictionary). case := OrderedDictionary newFromPairs: { [ cond ]. [ do ]. [ cond ]. [ do ]. [ true ]. [ otherwise do ] } case keysAndValuesDo: [ :cond :do | (cond value: expression) ifTrue: [ “you need to break here" ^ do value ]. 2) To use a “case method” caseMethod: expression expression = cond1 ifTrue: [ ^ do1 ]. expression = cond2 ifTrue: [ ^ do2 ]. ^ otherwise Personally, I prefer the second one because is more concise, performant and easy to understand. Esteban

On 28 Dec 2019, at 11:58, Dennis Schetinin <chaetal@gmail.com> wrote:

It would be an overkill to do it for this particular case, but Smalltalk makes it possible to implement a case-like construction:

[ expression ] when: [ :value | condition1 ] do: [-0do :value | ... ]; when: [ :value | condition2 ] do: [ :value | ... ]; otherwiseDo: [ :value | ... ]; evaluate

I am sure, something like this has been implemented already somewhere (maybe in Squeak?). Still not sure it is practical as compared to simple if-s, and for sure not widely used :) ...On the other hand, sometimes the case-like construction can be considered a more intension-revealing style.

пт, 27 дек. 2019 г., 22:18 Roelof Wobben via Pharo-users <pharo-users@lists.pharo.org <mailto:pharo-users@lists.pharo.org>>: Hello,

Im trying to solve a challenge from exercism where I have to calculate the points somehow gets on a very simple darts board.

I solved it like this :

scoreX: anInteger y: anInteger2 | distance | distance := (anInteger squared + anInteger2 squared) sqrt. distance > 10 ifTrue: [ ^ 0 ]. distance > 5 ifTrue: [ ^ 1 ]. distance > 1 ifTrue: [ ^ 5 ]. ^ 10

but now I use three if then and I think it's ugly code.

Is there a way I can make it more the smalltalk way ?

Regards,

Roelof