我想询问是否可以通过引用传递脚本函数的参数：
即在C中做一些这样的事情：
Void boo (int & myint) { myint= 5; }
main (){
int t= 4;
printf t; // t->4
boo (t);
printf t; // t ->5
}
那么在BASH我想做一些像：
Function boo ()
{
var1=$1 # now var1 is global to the scrip but using it outside
# this function makes me loose encapsulation
local var2=$1 # so i should use a local variable ... but how to pass it back ?
var2='new' # only changes the local copy
#$1='new' this is wrong of course ...
# ${!1}='new' # can i somehow use indirect reference?
}
# call boo
SOME_VAR='old'
echo $SOME_VAR # -> old
boo "$SOME_VAR"
echo $SOME_VAR # -> new
任何想法都不胜感激。
使用助手功能upvar：
# Assign variable one scope above the caller.
# Usage: local "$1" && upvar $1 value [value ...]
# Param: $1 Variable name to assign value to
# Param: $* Value(s) to assign. If multiple values,an array is
# assigned,otherwise a single value is assigned.
# NOTE: For assigning multiple variables,use 'upvars'. Do NOT
# use multiple 'upvar' calls,since one 'upvar' call might
# reassign a variable to be used by another 'upvar' call.
# See: http://fvue.nl/wiki/Bash:_Passing_variables_by_reference
upvar() {
if unset -v "$1"; then # Unset & validate varname
if (( $# == 2 )); then
eval $1=\"\$2\" # Return single value
else
eval $1=\(\"\${@:2}\"\) # Return array
fi
fi
}
并在Newfun()中使用它：
local "$1" && upvar $1 new
为了返回多个变量，请使用另一个帮助函数upvars。这允许在一个调用中传递多个变量，从而避免如果一个upvar调用更改另一个后续upvar调用中使用的变量的可能冲突。
请参阅：http://www.fvue.nl/wiki/Bash:_Passing_variables_by_reference for helper function upvars和更多信息。
问题在于：
eval $1=new
是否如果$ 1碰巧包含一个命令是不安全的：
set -- 'ls /;true' eval $1=new # Oops
最好使用printf -v：
printf -v "$1" %s new
但是printf -v不能分配数组。
此外，如果变量恰好被声明为local，那么eval和printf都不会工作：
g() { local b; eval $1=bar; } # WRONG
g b # Conflicts with `local b'
echo $b # b is empty unexpected
即使本地b未设置，冲突仍然存在：
g() { local b; unset b; eval $1=bar; } # WRONG
g b # Still conflicts with `local b'
echo $b # b is empty unexpected
