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以下问题的解决方案是否正确?什么是合法的erlang警卫?
从列表中选择合法的警戒表达式,变量A已经绑定。
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)
我的解决方案(正确卫士):
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)