Is there a way to use local function in guard

2020-04-02 07:46发布

Here is my code:

is_prime(Num)->
    length(list_of_dividers(Num)) == 0.

problem_7(Current, Primes, Counter) when Primes >= 10001->
    Current;
problem_7(Current, Primes, Counter) when is_prime(Counter) ->
    problem_7(Counter, Primes + 1, Counter + 1);
problem_7(Current, Primes, Counter) ->
    problem_7(Current, Primes, Counter).

I get error:

32> c(problem_7).
./problem_7.erl:30: call to local/imported function is_prime/1 is illegal in guard

And i can not use local function in the 'if'-expression:

if is_prime(Counter)->
    problem_7(Counter, Primes + 1, Counter + 1);
   true ->
    problem_7(Current, Primes, Counter + 1)
end.

I am just wondering if there is any way to use local function in guard and how can I write condition with local function?

标签: erlang
1条回答
Anthone
2楼-- · 2020-04-02 08:36

The reason for restricting the set of valid expressions is that evaluation of a guard expression must be guaranteed to be free of side effects. http://www.erlang.org/doc/reference_manual/expressions.html (section 7.24)

Use case statement inside your function. You should be able to use local function in if and case.

Edited: Agree with @cthulahoops, I was wrong about if http://www.erlang.org/doc/reference_manual/expressions.html#id75927

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