当我有一个函数本地只读变量和具有相同名称的全局只读变量时,我会得到一个惊人的行为.
当从全局声明中删除只读选项时.即
declare -r var="main"
改为：
declare var="main"
我得到了预期的行为.我一直在阅读bash手册页,但我找不到这个行为的解释.你能请我指出手册中解释这个问题的部分吗？
我认为这是一个类似的问题,比How does lexical scoping supported in different shell languages?更具体.
细节：
$cat readonly_variable.sh
#!/bin/bash
# expected output:
#
# BASH_VERSION = 3.2.25(1)-release
# function
# main
#
# but instead getting:
#
# BASH_VERSION = 3.2.25(1)-release
# ./readonly_variable.sh: line 6: local: var: readonly variable
# main
# main
#
# when read-only option (-r) is removed from global declaration (*),the output
# is expected
set -o nounset
function func {
local -r var="function"
echo "$var"
}
declare -r var="main" # (*)
echo BASH_VERSION = $BASH_VERSION
echo $(func)
echo $var
exit 0
我陷入了这个特殊的Bash版本.
$./readonly_variable.sh BASH_VERSION = 3.2.25(1)-release ./readonly_variable.sh: line 24: local: var: readonly variable main main $
实际上,为了安全起见,明确地禁止只读全局变量的本地副本,如bash源代码(在variables.c：make_local_variable中)所示：
The test against old_var’s context level is to disallow local copies of readonly global variables (since “I” believe that this could be a security hole).
(其中“我”不是我,我只是引用)
/* Since this is called only from the local/declare/typeset code,we can
call builtin_error here without worry (of course,it will also work
for anything that sets this_command_name). Variables with the `noassign'
attribute may not be made local. The test against old_var's context
level is to disallow local copies of readonly global variables (since I
believe that this could be a security hole). Readonly copies of calling
function local variables are OK. */
if (old_var && (noassign_p (old_var) ||
(readonly_p (old_var) && old_var->context == 0)))
{
if (readonly_p (old_var))
sh_readonly (name);
return ((SHELL_VAR *)NULL);
}
