ksh 和 bash 脚本之间的区别

考虑以下脚本（用于 local_var2 的算术语法与这种情况无关）：

#!/bin/ksh

function my_func1
{
   typeset local_var1=one
   typeset local_var2
   (( local_var2 = 1 ))
   echo my_func1: local_var1 = $local_var1, local_var2 = $local_var2
}

my_func2()
{
   typeset local_var1=two
   typeset local_var2
   (( local_var2 = 2 ))
   echo my_func2: local_var1 = $local_var1, local_var2 = $local_var2
}
local_var1=0
local_var2=0
echo before functions: local_var1 = $local_var1, local_var2 = $local_var2

my_func1
echo after my_func1: local_var1 = $local_var1, local_var2 = $local_var2

my_func2
echo after my_func2: local_var1 = $local_var1, local_var2 = $local_var2

运行时会产生以下输出：

before functions: local_var1 = 0, local_var2 = 0
my_func1: local_var1 = one, local_var2 = 1
after my_func1: local_var1 = 0, local_var2 = 0
my_func2: local_var1 = two, local_var2 = 2
after my_func2: local_var1 = two, local_var2 = 2

（这不是预期的！）

如果我在 bash 中运行相同的脚本，输出为：

before functions: local_var1 = 0, local_var2 = 0
my_func1: local_var1 = one, local_var2 = 1
after my_func1: local_var1 = 0, local_var2 = 0
my_func2: local_var1 = two, local_var2 = 2
after my_func2: local_var1 = 0, local_var2 = 0

（这是所期望的！）

这是怪异之处之一ksh93.

typeset将变量定义为局部范围的变量仅适用于函数定义样式：

function func_name
{
}

不符合函数定义风格：

func_name()
{
}

随着func_name()风格，一切都是全球性的。所以行为ksh正如预期的那样！

But bash显然比ksh在这方面。因此，当以下情况时，它将两个函数中的变量范围设置为局部变量：typeset被使用了。

There's FAQ http://kornshell.com/doc/faq.htmlksh 文档中的条目说明了函数定义之间的差异：

Q18. What is the difference between function name and name()?

A18. In ksh88 these were the same.  However, the POSIX standard
    choose foo() for functions and defined System V Release 2
    semantics to them so that there are no local variables
    and so that traps are not scoped.  ksh93 keeps the ksh88
    semantics for functions defined as function name, and
    has changed the name() semantics to match the POSIX
    semantics.  Clearly, function name is more useful.