我愿提出一个论点,在清单产生另一个清单时,这一论点会取得成功。
例如?-proceed_list([1,2],] /。 是的还是真实的(无论汇编者是什么)。
Can anyone help me?
你的问题很难说。 正如Aqua所评论的那样,你应当重新措辞。 关于其价值,我理解你的例子,即你想检查第二个名单(> 是第一个<代码>[1,2](即“继续”意义上的“收到”)的尾声。 如果这是你想要的,那么这项工作应当做到:
proceed_list(L1, L2) :-
once(append(_, L2, L1)).
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