(a + 1) 不一定 > a
by 高顯忠, 2010-12-02 22:03, Views(1881)
| 隱藏詳細資料 21:16 (45 分鐘前) |
program VF0904
implicit none
integer a, b
a= 10
b= a + 1
! b= 11, > 10, b > a
print *, 'a, b= ', a, b
pause
! -----------------------------------------------------
do while (b .GT. a)
a= a + 1
b= a + 1
end do
! b <= a, 怎麼可能 (a + 1) <= a
print *, 'exit do while(), b <= a'
print *, 'a, b= ', a, b
pause
end program VF0904
! -----------------------------------------------------
!dec$if(.false.)
a, b= 10 11
Fortran Pause - Enter command<CR> or <CR> to continue.
exit do while(), b <= a
a, b= 2147483647 -2147483648
Fortran Pause - Enter command<CR> or <CR> to continue.
!dec$endif
implicit none
integer a, b
a= 10
b= a + 1
! b= 11, > 10, b > a
print *, 'a, b= ', a, b
pause
! ------------------------------
do while (b .GT. a)
a= a + 1
b= a + 1
end do
! b <= a, 怎麼可能 (a + 1) <= a
print *, 'exit do while(), b <= a'
print *, 'a, b= ', a, b
pause
end program VF0904
! ------------------------------
!dec$if(.false.)
a, b= 10 11
Fortran Pause - Enter command<CR> or <CR> to continue.
exit do while(), b <= a
a, b= 2147483647 -2147483648
Fortran Pause - Enter command<CR> or <CR> to continue.
!dec$endif