Mar*_*ers 8

  • foo[x, n]返回foo从第x 个元素开始的n个元素.
  • foo[x][y]找到第x个元素foo并返回其第y个元素.

这是一个示例:

> foo = [[0,1,2],[3,4,5],[6,7,8],[9,10,11],[12,13,14]]
=> [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10, 11], [12, 13, 14]]
> foo[1,2]
=> [[3, 4, 5], [6, 7, 8]]
> foo[1][2]
=> 5
Run Code Online (Sandbox Code Playgroud)