在制作 Connect Four 游戏时,我在一个名为 的函数中遇到了一个奇怪的问题make_move,当两个等效的 return 语句表现不同时。
唯一的直接依赖函数是put_piece(board, column, player),它将玩家的棋子放在棋盘给定列中最底部的空白位置。put_piece返回一个包含两个元素的元组:棋子结束的行的索引(如果列已满,则为 -1)和更新后的棋盘。该put_piece功能已正确实现。
该make_move功能是在分歧发生。如果我使用通常的 if else 返回表示法实现,它会成功返回row(放置棋子的行的索引)和board(更新的板),如下所示:
def make_move(board, max_rows, max_cols, col, player):
"""
Put player's piece in column COL of the board, if it is a valid move.
Return a tuple of two values:
1. If the move is valid, make_move returns the index of the row the
piece is placed in. Otherwise, it returns -1.
2. The updated board
"""
if 0 <= col < len(board[0]):
return put_piece(board, max_rows, col, player)
return -1, board
这是make_move应该如何返回:
>>> rows, columns = 2, 2
>>> board = create_board(rows, columns)
>>> row, board = make_move(board, rows, columns, 0, 'X')
>>> row
1
>>> board
[['-', '-'], ['X', '-']]
但是,如果我更改make_move为
def make_move(board, max_rows, max_cols, col, player):
"""
Put player's piece in column COL of the board, if it is a valid move.
Return a tuple of two values:
1. If the move is valid, make_move returns the index of the row the
piece is placed in. Otherwise, it returns -1.
2. The updated board
"""
return put_piece(board, max_rows, col, player) if 0 <= col < len(board[0]) else -1, board
两个返回值都作为一个元组分配给row,并board简单地继承前一个值。
>>> rows, columns = 2, 2
>>> board = create_board(rows, columns)
>>> row, board = make_move(board, rows, columns, 0, 'X')
>>> row
(1, [['-', '-'], ['X', '-']])
>>> board
[['-', '-'], ['-', '-']]
除了符号之外,两种编写函数的方式在字面上是相同的。知道为什么会这样吗?
桃花长相依
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