巴什 - 按引用传递参数 [英] Bash - Passing arguments by reference

问题描述

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.

推荐答案

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 ...

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