Question

我有一个永不停止执行的fortran程序。

program russianmultiplication
implicit none
integer::x,y,ginx,giny,k=0

print*,'give a number'
read*,x
k=k+x
print*,'one more'
read*,y
if (y==1) then
    print*,x
end if                
do while(y/=1)
    ginx=x*2
    giny=y/2
    if (mod(giny,2)/=0) then
        k=k+ginx
    end if
end do
print*,'result',k
end program

为什么这个程序永远不会结束？

Answer 1

你永远不会在循环中更改y，也没有exit语句。因此，y在输入时是1，在这种情况下，循环的条件永远不会为真且永远不会执行，或者您永远不会离开循环，因为条件始终为真。

Answer 2

这可能是原始程序Russian peasant multiplication的更清晰版本：

PROGRAM Russian_peasant_multiplication
IMPLICIT NONE
INTEGER :: total, x, x_save, y, y_save

total = 0!running total
PRINT *, "x ="
READ *, x
x_save = x
total = total + x
PRINT *, "y ="
READ *, y
y_save = y
PRINT *, y, x
!do while(y/=1)
modulo: DO
  IF (y == 1) EXIT!replace DO WHILE for better control
  !ginx=x*2
  x = x * 2!okay to redefine since saved
  !giny=y/2
  y = y / 2!okay to redefine since saved
  !if (mod(giny,2)/=0) then
  IF (MOD(y, 2) /= 0) THEN!keep only odd numbers on the left
    PRINT *, y, x
    !k=k+ginx
    total = total + x
  END IF
END DO modulo
PRINT *, " "
PRINT *, x_save
PRINT *, "x"
PRINT *, y_save
PRINT *, "="
PRINT *, total
END PROGRAM Russian_peasant_multiplication

我添加了有用的评论，并指出逻辑存在缺陷的地方。希望这会有所帮助。

我的fortran程序永远不会停止执行

2 个答案: