做 loop/if 语句逻辑

Question

我正在开发一个井字游戏程序。函数 CHECK_WINNER 应该接收给定状态下的棋盘，并确定游戏是否获胜、平局，或者玩家是否需要继续游戏。 CHECK_WINNER 根据棋盘的状态取不同的值。

win(n;1:3) 块列出了井字棋盘的 8 种可能获胜配置。

我认为问题出在 do 循环中。我的目标是为每个玩家（1 和 2）循环遍历棋盘的每个获胜场景，并检查是否满足获胜条件。 CHECK_MOVE 如果玩家 1 获胜则应为 1，如果玩家 2 获胜则应为 2，如果游戏平局则应为 3，对于任何其他情况应为 0。我究竟做错了什么？

program checkwinner
implicit none
integer, dimension(9) :: board
integer, external :: CHECK_WINNER
integer :: cw

!board = (/ 1,2,1,2,1,2,2,2,1 /)  ! not working correctly. 1 is winner, board full; cw returns 3
!board = (/ 1,2,1,2,0,2,2,2,1 /) ! working correctly. no winner, open spaces; cw returns 0
!board = (/ 1,1,1,0,0,0,0,0,0 /) ! not working. cw should return 1; instead cw returns 0
board = (/ 2,2,2,0,0,0,0,0,0 /) ! not working

cw = CHECK_WINNER(board)
print *, board(1:3)
print *, board(4:6)
print *, board(7:9)
print *, cw

end program checkwinner
!!!!!!!!!!!!!!!!!!!!!!!!!
!!!!!!!!!!!!!!!!!!!!!!!!!
integer function CHECK_WINNER(fboard)
implicit none
integer :: i
integer, dimension(8,3) :: win
integer, dimension(9), intent(in) :: fboard

win(1,1:3) = (/ 1,2,3 /)
win(2,1:3) = (/ 4,5,6 /)
win(3,1:3) = (/ 7,8,9 /)
win(4,1:3) = (/ 1,4,7 /)
win(5,1:3) = (/ 2,5,8 /)
win(6,1:3) = (/ 3,6,9 /)
win(7,1:3) = (/ 1,5,9 /)
win(8,1:3) = (/ 3,5,7 /)

do i = 1,8
     if (fboard(win(i,1)) == 1 .and. fboard(win(i,2)) == 1 .and. fboard(win(i,3)) == 1) then
        CHECK_WINNER = 1
    else if (fboard(win(i,1)) == 2 .and. fboard(win(i,2)) == 2 .and. fboard(win(i,3)) == 2) then
        CHECK_WINNER = 2
    else if ((fboard(win(i,1)) == 1 .or. fboard(win(i,1)) == 2) .and. (fboard(win(i,2)) == 1 .or. &
        fboard(win(i,2)) == 2) .and. (fboard(win(i,3)) == 1 .or. fboard(win(i,3)) == 2)) then
        CHECK_WINNER = 3
    else
        CHECK_WINNER = 0
    end if
end do

end function CHECK_WINNER

Answer 1

因此，在您在代码中使用 board = (/ 2,2,2,0,0,0,0,0,0 /) 给出的示例中，当函数的 do 循环中 i=1 时，它正确设置了 CHECK_WINNER=2 因为

fboard(win(i,1)) == 2 .and. fboard(win(i,2)) == 2 .and. fboard(win(i,3)) == 2

是真的。不幸的是，当循环到达 i=2、i=3 等时，它会将 CHECK_WINNER 重置为零，因为获胜条件的 none 匹配。这是导致问题的 else 块。将 do 循环更改为

CHECK_WINNER = 0

do i = 1,8
  if (fboard(win(i,1)) == 1 .and. fboard(win(i,2)) == 1 .and. fboard(win(i,3)) == 1) then
    CHECK_WINNER = 1
  else if (fboard(win(i,1)) == 2 .and. fboard(win(i,2)) == 2 .and. fboard(win(i,3)) == 2) then
    CHECK_WINNER = 2
  else if ((fboard(win(i,1)) == 1 .or. fboard(win(i,1)) == 2) .and. (fboard(win(i,2)) == 1 .or. &
    fboard(win(i,2)) == 2) .and. (fboard(win(i,3)) == 1 .or. fboard(win(i,3)) == 2)) then
    CHECK_WINNER = 3
   end if
end do

因此，如果它找到已知组合，它会设置 CHECK_WINNER。您也可以在每个 (else)if 子句中使用 exit 找到已知组合后立即退出循环。（设置CHECK_WINNER后）

做 loop/if 语句逻辑

do loop/if statement logic

fortran

fortran90

do-loops