巴什 - 按引用传递参数
it1352
帮助1人
问题说明
我想问是否可以通过引用参数传递给一个脚本函数：
I want to ask if it is possible to pass arguments to a script function by reference:
即。做一些看起来像在C：
i.e. to do something that would look like this in C:
Void boo (int & myint) { myint= 5; }
main (){
int t= 4;
printf t; // t->4
boo (t);
printf t; // t ->5
}
因此，在BASH然后，我想做的事是这样的：
So then in BASH I want to do something like:
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，将AP preciated。
Any thoughts would be appreciated.
正确答案
#1
我已经找到一种方法来做到这一点，但我不知道如何正确的是这样的：
I have found a way to do this but I am not sure how correct this is:
Newfun()
{
local var1="$1"
eval $var1=2
# or can do eval $1=2 if no local var
}
var=1
echo var is $var # $var = 1
newfun 'var' # pass the name of the variable…
echo now var is $var # $var = 2
所以我们传递，而不是值，然后在变量名使用eval ...
So we pass the variable name as opposed to the value and then use eval ...
这篇好文章是转载于：学新通技术网
