CODEWITHSUNDEEP
listif-statementerlang
beoliver
beoliver

Reputation: 5759

cons erlang if stament into list

is it possible to us an if statement in a similar way to Haskell in the following structure?

replace_with :: (a -> Bool) -> [a] -> a -> [a]
replace_with _ [] _ = []
replace_with f (x:xs) y = (if f x then y else x) : replace_with f xs y

I have tried the following:

rwith(_,[],_) ->
    [];
rwith(F,[X|XS],Y) ->
    [if F(X) == true -> Y; true -> X, end. | replace_with(F,XS,Y)].

but I get the compile response: 

listsh.erl:40: syntax error before: 'end'
listsh.erl:40: syntax error before: '|'

I have used the following which works fine, but am interested in knowing if consing the return of an if statement is possible?

replace_with(_,[],_) ->
    [];
replace_with(F,[X|XS],Y) ->
    case F(X) of
        true -> [Y|replace_with(F,XS,Y)];
        false -> [X|replace_with(F,XS,Y)]
    end.

Upvotes: 1

Views: 75

Answers (1)

Hynek -Pichi- Vychodil
Hynek -Pichi- Vychodil

Reputation: 26121

Correct code should be:

[if F(X) =:= true -> Y; true -> X end | replace_with(F,XS,Y)].

(using =:= is more convenient in Erlang but doesn't matter here) but there can be only guard expression in if statement so you have to use 

T = F(X),
[if T =:= true -> Y; true -> X end | replace_with(F,XS,Y)].

or more idiomatic

[case F(X) of true -> Y; false -> X end | replace_with(F,XS,Y)].

You can even replace the whole function with

[ case F(X) of true -> Y; false -> X end || X <- XS ].

Note Erlang syntax here, , is separator of expressions, ; is separator of clauses, . ends the function definition, end ends compound expression. It can be inconvenient for someone who comes from other languages.

Upvotes: 4

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