第 34 章. 陷阱

以下是一些（不推荐！）的脚本编写实践，它们会给原本枯燥的生活带来刺激。

• 将保留字或字符分配给变量名。

  case=value0 # Causes problems. 23skidoo=value1 # Also problems. # Variable names starting with a digit are reserved by the shell. # Try _23skidoo=value1. Starting variables with an underscore is okay. # However . . . using just an underscore will not work. _=25 echo $_ # $_ is a special variable set to last arg of last command. # But . . . _ is a valid function name! xyz((!*=value2 # Causes severe problems. # As of version 3 of Bash, periods are not allowed within variable names.

• 在变量名（或函数名）中使用连字符或其他保留字符。

  var-1=23 # Use 'var_1' instead. function-whatever () # Error # Use 'function_whatever ()' instead. # As of version 3 of Bash, periods are not allowed within function names. function.whatever () # Error # Use 'functionWhatever ()' instead.

• 变量和函数使用相同的名称。这会使脚本难以理解。

  do_something () { echo "This function does something with \"$1\"." } do_something=do_something do_something do_something # All this is legal, but highly confusing.

• 不恰当地使用空白字符。与其他编程语言相比，Bash 对空白字符可能非常挑剔。

  var1 = 23 # 'var1=23' is correct. # On line above, Bash attempts to execute command "var1" # with the arguments "=" and "23". let c = $a - $b # Instead: let c=$a-$b or let "c = $a - $b" if [ $a -le 5] # if [ $a -le 5 ] is correct. # ^^ if [ "$a" -le 5 ] is even better. # [[ $a -le 5 ]] also works.

• 在花括号内的代码块中，最后一个命令没有用分号结尾。

  { ls -l; df; echo "Done." } # bash: syntax error: unexpected end of file { ls -l; df; echo "Done."; } # ^ ### Final command needs semicolon.

• 假设未初始化的变量（在赋值之前的变量）会被“置零”。未初始化的变量的值为null，而不是零。

  #!/bin/bash echo "uninitialized_var = $uninitialized_var" # uninitialized_var = # However . . . # if $BASH_VERSION ≥ 4.2; then if [[ ! -v uninitialized_var ]] then uninitialized_var=0 # Initialize it to zero! fi

• 在测试中混淆 = 和 -eq。记住，= 用于比较字面变量，而 -eq 用于整数。

  if [ "$a" = 273 ] # Is $a an integer or string? if [ "$a" -eq 273 ] # If $a is an integer. # Sometimes you can interchange -eq and = without adverse consequences. # However . . . a=273.0 # Not an integer. if [ "$a" = 273 ] then echo "Comparison works." else echo "Comparison does not work." fi # Comparison does not work. # Same with a=" 273" and a="0273". # Likewise, problems trying to use "-eq" with non-integer values. if [ "$a" -eq 273.0 ] then echo "a = $a" fi # Aborts with an error message. # test.sh: [: 273.0: integer expression expected

• 错误使用字符串比较运算符。

  示例 34-1. 数值比较和字符串比较不等价

  #!/bin/bash # bad-op.sh: Trying to use a string comparison on integers. echo number=1 # The following while-loop has two errors: #+ one blatant, and the other subtle. while [ "$number" < 5 ] # Wrong! Should be: while [ "$number" -lt 5 ] do echo -n "$number " let "number += 1" done # Attempt to run this bombs with the error message: #+ bad-op.sh: line 10: 5: No such file or directory # Within single brackets, "<" must be escaped, #+ and even then, it's still wrong for comparing integers. echo "---------------------" while [ "$number" \< 5 ] # 1 2 3 4 do # echo -n "$number " # It *seems* to work, but . . . let "number += 1" #+ it actually does an ASCII comparison, done #+ rather than a numerical one. echo; echo "---------------------" # This can cause problems. For example: lesser=5 greater=105 if [ "$greater" \< "$lesser" ] then echo "$greater is less than $lesser" fi # 105 is less than 5 # In fact, "105" actually is less than "5" #+ in a string comparison (ASCII sort order). echo exit 0

• 尝试使用 let 来设置字符串变量。

  let "a = hello, you" echo "$a" # 0

• 有时，“test” 方括号 ([ ]) 中的变量需要被引用（双引号）。不这样做可能会导致意外行为。请参阅示例 7-6、示例 20-5 和 示例 9-6。

• 引用包含空白字符的变量会阻止分割。有时这会产生意想不到的后果。

• 从脚本发出的命令可能无法执行，因为脚本所有者缺乏执行它们的权限。如果用户无法从命令行调用命令，那么将其放入脚本中同样会失败。尝试更改相关命令的属性，甚至可以设置 suid 位（当然，以 root 身份）。

• 尝试使用 - 作为重定向运算符（它不是）通常会导致不愉快的意外。

  command1 2> - | command2 # Trying to redirect error output of command1 into a pipe . . . # . . . will not work. command1 2>& - | command2 # Also futile. Thanks, S.C.

• 使用 Bash 版本 2+ 的功能可能会导致因错误消息而中止。较旧的 Linux 机器可能默认安装了 Bash 1.XX 版本。

  #!/bin/bash minimum_version=2 # Since Chet Ramey is constantly adding features to Bash, # you may set $minimum_version to 2.XX, 3.XX, or whatever is appropriate. E_BAD_VERSION=80 if [ "$BASH_VERSION" \< "$minimum_version" ] then echo "This script works only with Bash, version $minimum or greater." echo "Upgrade strongly recommended." exit $E_BAD_VERSION fi ...

• 在 Bourne shell 脚本中使用 Bash 特有的功能（#!/bin/sh）在非 Linux 机器上 可能会导致意外行为。Linux 系统通常将 sh 别名为 bash，但这对于通用的 UNIX 机器不一定成立。

• 在 Bash 中使用未记录的功能被证明是一种危险的做法。本书的先前版本中有几个脚本依赖于“特性”，即虽然 exit 或 return 值的最大值为 255，但该限制不适用于负整数。不幸的是，在 2.05b 及更高版本中，该漏洞消失了。请参阅 示例 24-9。

• 在某些情况下，可能会返回误导性的 退出状态。这可能发生在在函数中设置局部变量或将算术值赋给变量时。

• 算术表达式的退出状态不等同于错误代码。

  var=1 && ((--var)) && echo $var # ^^^^^^^^^ Here the and-list terminates with exit status 1. # $var doesn't echo! echo $? # 1

• 带有 DOS 类型换行符的脚本（\r\n）将无法执行，因为#!/bin/bash\r\n是不被识别的，不同于预期的#!/bin/bash\n。解决方法是将脚本转换为 UNIX 样式的换行符。

  #!/bin/bash echo "Here" unix2dos $0 # Script changes itself to DOS format. chmod 755 $0 # Change back to execute permission. # The 'unix2dos' command removes execute permission. ./$0 # Script tries to run itself again. # But it won't work as a DOS file. echo "There" exit 0

• 以#!/bin/sh开头的 shell 脚本将不会以完整的 Bash 兼容模式运行。某些 Bash 特有的功能可能会被禁用。需要完全访问所有 Bash 特有扩展的脚本应该以#!/bin/bash.

• 在 here 文档的终止限制字符串前面放置空白字符将导致脚本中出现意外行为。

• 在一个输出被捕获的函数中放置多个 echo 语句。

  add2 () { echo "Whatever ... " # Delete this line! let "retval = $1 + $2" echo $retval } num1=12 num2=43 echo "Sum of $num1 and $num2 = $(add2 $num1 $num2)" # Sum of 12 and 43 = Whatever ... # 55 # The "echoes" concatenate.

  这将不起作用。

• 脚本可能无法将变量 export 回其父进程、shell 或环境。正如我们在生物学中学到的，子进程可以从父进程继承，但反之则不然。

  WHATEVER=/home/bozo export WHATEVER exit 0

  bash$ echo $WHATEVER bash$

  果然，回到命令提示符下，$WHATEVER 仍然未设置。

• 在子 shell 中设置和操作变量，然后尝试在子 shell 范围之外使用这些相同的变量将导致不愉快的意外。

  示例 34-2. 子 Shell 的陷阱

  #!/bin/bash # Pitfalls of variables in a subshell. outer_variable=outer echo echo "outer_variable = $outer_variable" echo ( # Begin subshell echo "outer_variable inside subshell = $outer_variable" inner_variable=inner # Set echo "inner_variable inside subshell = $inner_variable" outer_variable=inner # Will value change globally? echo "outer_variable inside subshell = $outer_variable" # Will 'exporting' make a difference? # export inner_variable # export outer_variable # Try it and see. # End subshell ) echo echo "inner_variable outside subshell = $inner_variable" # Unset. echo "outer_variable outside subshell = $outer_variable" # Unchanged. echo exit 0 # What happens if you uncomment lines 19 and 20? # Does it make a difference?

• 将 echo 输出管道传输到 read 可能会产生意外结果。在这种情况下，read 的行为就像它在子 shell 中运行一样。相反，请使用 set 命令（如 示例 15-18 中所示）。

  示例 34-3. 将 echo 的输出管道传输到 read

  #!/bin/bash # badread.sh: # Attempting to use 'echo and 'read' #+ to assign variables non-interactively. # shopt -s lastpipe a=aaa b=bbb c=ccc echo "one two three" | read a b c # Try to reassign a, b, and c. echo echo "a = $a" # a = aaa echo "b = $b" # b = bbb echo "c = $c" # c = ccc # Reassignment failed. ### However . . . ## Uncommenting line 6: # shopt -s lastpipe ##+ fixes the problem! ### This is a new feature in Bash, version 4.2. # ------------------------------ # Try the following alternative. var=`echo "one two three"` set -- $var a=$1; b=$2; c=$3 echo "-------" echo "a = $a" # a = one echo "b = $b" # b = two echo "c = $c" # c = three # Reassignment succeeded. # ------------------------------ # Note also that an echo to a 'read' works within a subshell. # However, the value of the variable changes *only* within the subshell. a=aaa # Starting all over again. b=bbb c=ccc echo; echo echo "one two three" | ( read a b c; echo "Inside subshell: "; echo "a = $a"; echo "b = $b"; echo "c = $c" ) # a = one # b = two # c = three echo "-----------------" echo "Outside subshell: " echo "a = $a" # a = aaa echo "b = $b" # b = bbb echo "c = $c" # c = ccc echo exit 0

  事实上，正如 Anthony Richardson 指出的那样，管道传输到任何循环都可能导致类似的问题。

  # Loop piping troubles. # This example by Anthony Richardson, #+ with addendum by Wilbert Berendsen. foundone=false find $HOME -type f -atime +30 -size 100k | while true do read f echo "$f is over 100KB and has not been accessed in over 30 days" echo "Consider moving the file to archives." foundone=true # ------------------------------------ echo "Subshell level = $BASH_SUBSHELL" # Subshell level = 1 # Yes, we're inside a subshell. # ------------------------------------ done # foundone will always be false here since it is #+ set to true inside a subshell if [ $foundone = false ] then echo "No files need archiving." fi # =====================Now, here is the correct way:================= foundone=false for f in $(find $HOME -type f -atime +30 -size 100k) # No pipe here. do echo "$f is over 100KB and has not been accessed in over 30 days" echo "Consider moving the file to archives." foundone=true done if [ $foundone = false ] then echo "No files need archiving." fi # ==================And here is another alternative================== # Places the part of the script that reads the variables #+ within a code block, so they share the same subshell. # Thank you, W.B. find $HOME -type f -atime +30 -size 100k | { foundone=false while read f do echo "$f is over 100KB and has not been accessed in over 30 days" echo "Consider moving the file to archives." foundone=true done if ! $foundone then echo "No files need archiving." fi }

  当尝试写入以下内容的stdout的 tail -f 管道传输到 grep。

  tail -f /var/log/messages | grep "$ERROR_MSG" >> error.log # The "error.log" file will not have anything written to it. # As Samuli Kaipiainen points out, this results from grep #+ buffering its output. # The fix is to add the "--line-buffered" parameter to grep.

• 在脚本中使用 “suid” 命令是危险的，因为它可能会危及系统安全。[1]

• 使用 shell 脚本进行 CGI 编程可能会有问题。Shell 脚本变量不是 “类型安全” 的，这可能会导致与 CGI 相关的意外行为。此外，很难对 shell 脚本进行 “防破解”。

• Bash 不能正确处理 双斜杠 (//) 字符串。

• 为 Linux 或 BSD 系统编写的 Bash 脚本可能需要修复才能在商业 UNIX 机器上运行。此类脚本通常使用 GNU 命令和过滤器集，它们比通用的 UNIX 对等物具有更强大的功能。对于诸如 tr 之类的文本处理实用程序尤其如此。

• 遗憾的是，Bash 本身的更新破坏了以前可以完美运行的旧脚本。让我们回顾一下使用未记录的 Bash 功能有多么危险。