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以下問題的解決方案是否正確?什麼是合法的erlang警衛?
從列表中選擇合法的警戒表達式,變量A已經綁定。
true, false, apple, 1+2, 1+2 > 3, is_atom(A), B = 3, A = 3, A == 3,length(A), lists:max(A), list_to_atom(A), A and B, (A > 3) and (A < 12)
我的解決方案(正確衛士):
true, false, 1+2 > 3, is_atom(A), A == 3, length(A), lists:max(A), list_to_atom(A), A and B, (A > 3) and (A < 12)