Shell命令替换场景下函数返回值异常问题求助

我编写了一个可在终端指定位置打印内容的函数print_in_specified_position，还实现了一个用于更新并返回新打印位置的函数get_new_line，但后者通过命令替换调用时无法返回正确值，始终返回初始值。想咨询是否可以通过将该函数作为参数传递的方式让代码正常工作？

相关代码 #

function print_in_specified_position() {
  local X="$1"
  local Y="$2"
  local STRING="$3"
  tput cup "$X" "$Y"
  printf "$STRING"
}

function show_script_options() {
  COLUMS=$(tput cols)
  LINES=$(tput lines)
  export PRINT_X=$(( COLUMS / 2 - 77 ))
  export PRINT_Y=$(( LINES / 2 - 11 ))

  if (( PRINT_X < 0 )); then
    PRINT_X=0
  fi
  if (( PRINT_Y < 0 )); then
    PRINT_Y=0
  fi

  function get_new_line() {
    PRINT_Y=$(( PRINT_Y + 1 ))
    echo $PRINT_Y
  }

  print_in_specified_position "$PRINT_X" "$(get_new_line)" "\n →  What do you want to do?\n"
  print_in_specified_position "$PRINT_X" "$(get_new_line)" "\n    1 - Option 1"
  print_in_specified_position "$PRINT_X" "$(get_new_line)" "\n    2 - Option 2"
  print_in_specified_position "$PRINT_X" "$(get_new_line)" "\n    "
  print_in_specified_position "$PRINT_X" "$(get_new_line)" "\n    3 - Option 3"
  print_in_specified_position "$PRINT_X" "$(get_new_line)" "\n    4 - Option 4"
  print_in_specified_position "$PRINT_X" "$(get_new_line)" "\n    "
  print_in_specified_position "$PRINT_X" "$(get_new_line)" "\n    5 - Option 5"
  print_in_specified_position "$PRINT_X" "$(get_new_line)" "\n    "
  print_in_specified_position "$PRINT_X" "$(get_new_line)" "\n    6 - Option 6"
  print_in_specified_position "$PRINT_X" "$(get_new_line)" "\n    "
  print_in_specified_position "$PRINT_X" "$(get_new_line)" "\n    q - Quit\n"
}

测试情况 #

• 单独调用get_new_line时工作正常：

  get_new_line
  get_new_line
  get_new_line
  

  输出：

  14
  15
  16
  

• 使用命令替换调用时失效：

  echo "$(get_new_line)"
  echo "$(get_new_line)"
  echo "$(get_new_line)"
  

  输出：

  14
  14
  14
  

• 尝试"$get_new_line"和"${get_new_line}"写法，均无效。

问题原因及解决方法 #

问题根源 #

命令替换$(...)会创建子shell执行内部命令，子shell会继承父shell的变量，但子shell内的变量修改不会同步回父shell。所以每次$(get_new_line)调用时，PRINT_Y的修改只在当前子shell中生效，父shell的PRINT_Y始终保持初始值，导致每次返回的都是同一个数。

有效解决方法（无需传递函数参数） #

要让变量修改在当前shell生效，必须避免子shell。可以让get_new_line只负责更新变量，之后直接使用变量值：

function show_script_options() {
  COLUMS=$(tput cols)
  LINES=$(tput lines)
  PRINT_X=$(( COLUMS / 2 - 77 ))
  PRINT_Y=$(( LINES / 2 - 11 ))

  if (( PRINT_X < 0 )); then
    PRINT_X=0
  fi
  if (( PRINT_Y < 0 )); then
    PRINT_Y=0
  fi

  function get_new_line() {
    PRINT_Y=$(( PRINT_Y + 1 ))
  }

  get_new_line
  print_in_specified_position "$PRINT_X" "$PRINT_Y" "\n →  What do you want to do?\n"
  get_new_line
  print_in_specified_position "$PRINT_X" "$PRINT_Y" "\n    1 - Option 1"
  get_new_line
  print_in_specified_position "$PRINT_X" "$PRINT_Y" "\n    2 - Option 2"
  get_new_line
  print_in_specified_position "$PRINT_X" "$PRINT_Y" "\n    "
  get_new_line
  print_in_specified_position "$PRINT_X" "$PRINT_Y" "\n    3 - Option 3"
  get_new_line
  print_in_specified_position "$PRINT_X" "$PRINT_Y" "\n    4 - Option 4"
  get_new_line
  print_in_specified_position "$PRINT_X" "$PRINT_Y" "\n    "
  get_new_line
  print_in_specified_position "$PRINT_X" "$PRINT_Y" "\n    5 - Option 5"
  get_new_line
  print_in_specified_position "$PRINT_X" "$PRINT_Y" "\n    "
  get_new_line
  print_in_specified_position "$PRINT_X" "$PRINT_Y" "\n    6 - Option 6"
  get_new_line
  print_in_specified_position "$PRINT_X" "$PRINT_Y" "\n    "
  get_new_line
  print_in_specified_position "$PRINT_X" "$PRINT_Y" "\n    q - Quit\n"
}

这里不需要export变量，因为所有操作都在同一个shell函数内完成。

关于函数参数传递的可行性 #

Bash中可以传递函数名作为参数，但本质上还是调用函数，只要使用命令替换就会触发子shell，无法解决变量同步问题。所以传递函数参数不是解决该问题的有效方案，核心还是要避免子shell对变量修改的隔离。

内容的提问来源于stack exchange，提问作者João Bonfim