Question

我正在为正在执行的Battleships项目编写这段代码，我相信在将这些部分分配给子例程中的板子时遇到了逻辑错误。

“和（c ==（x + 1）或c ==（x-1））和（b ==（y + 1）或b ==（y-1））”

我相信我的错误在于此条件的这一部分，但我似乎看不到错误在哪里。我的代码错误可能不在其他地方，因为当用另一个测试条件替换该错误时，即IE的初始条件将放在子例程的第一部分：“如果c在range（0,3）和b在range（0， 3）”，可以根据需要绘制字符串“ X”

#FUNCTION - PLACE TILES
grid = [["  ","  ","  "],["  ","  ","  "],["  ","  ","  "]]
mrk = "X"
cruiser = 2

while True:
    try:
        x,y = input("which tile would you like to place your ship on?").split(",")
        x = int(x)
        y = int(y)
        if x in range(0,3) and y in range(0,3):
            grid[x][y] = mrk
            cruiser -= 1
        else:
            x,y = input("which tile would you like to place your ship on?").split(",")
        break
    except ValueError:
        x,y = input("INVALID CHARACTER\n\nwhich tile would you like to place your ship on?").split(",")

while cruiser < 2 and cruiser > 0:
    while True:
        try:
            c,b = input("which tile would you like to place your ship on?").split(",")
            c = int(c)
            b = int(b)
            print(c,b,x,y)
            if c in range(0,3) and b in range(0,3) and (c == (x+1) or c == (x-1)) and (b == (y+1) or b == (y-1)) :
                grid[c][b] = mrk
                cruiser -= 1

            else:
                c,b = input("INVALID POSITION\n\nwhich tile would you like to place your ship on?").split(",")
            break
        except ValueError:
            c,b = input("INVALID CHARACTER\n\nwhich tile would you like to place your ship on?").split(",")



print (grid[0][0]," | ",grid[0][1]," | ",grid[0][2])
print (grid[1][0]," | ",grid[1][1]," | ",grid[1][2])
print (grid[2][0]," | ",grid[2][1]," | ",grid[2][2])

由于它使用try和except，因此不会产生任何错误消息-但是，确实出现了我的代码中所示的INVALID POSITION。它应该接受该值，然后检查它是否仅与另一个点水平或垂直对齐，然后绘制它。

编辑：HTML / CSS代码段让您直观地看到我正在尝试做什么，谢谢

.grid-container {
  display: grid;
  grid-template: auto / auto auto auto;
  grid-gap: 5px;
  background-color: #000000;
  padding: 0px;
}

.grid-container>div {
  background-color: rgba(255, 255, 255, 0.8);
  text-align: center;
  padding: 20px 0;
  font-size: 30px;
}

<!DOCTYPE html>
<html>

<body>


  <p1> If the top right X was the first value plotted, valid positions for X would be everything marked X1, and invalid positions would be marked Y </p1>
  <div class="grid-container">
    <div class="item1">X</div>
    <div class="item2">X1</div>
    <div class="item3">X1</div>
    <div class="item4">X1</div>
    <div class="item5">Y</div>
    <div class="item6">Y</div>
    <div class="item7">X1</div>
    <div class="item8">Y</div>
    <div class="item9">Y</div>
  </div>

</body>

</html>

Answer 1

不确定我是否理解您想要做什么，但是我将替换掉

and (c == (x+1) or c == (x-1)) and (b == (y+1) or b == (y-1))

通过

and c != x and b != y

在我看来，您只想确认c和b与x和y不在同一位置。

更新：有了新的信息，您实际上希望将每个新标记放置在与前一个标记相邻的位置，而不是对角线，那么您确实应该存储最后一个标记的位置（例如prev_x，prev_y ）并在每个有效输入处进行更新，因此您可以将其与下一个输入的新值进行比较，依此类推。

除此之外，为防止将标记放置在对角线上，我认为没有魔术公式。您需要检查c和b是否不在对角线上。

[...]
    # avoid diagonals in relation to the previous marker
    and not (c == prev_x - 1 and b == prev_y - 1)
    and not (c == prev_x - 1 and b == prev_y + 1)
    and not (c == prev_x + 1 and b == prev_y - 1)
    and not (c == prev_x + 1 and b == prev_y + 1)
[...]

逻辑错误？ Python错误捕获和变量

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