shell declare variable using function argument

Question

I have the following code in shell

#!/bin/sh

function myfunc
{
   $1=3
   echo "myvar = $myvar"
}

myfunc myvar

expected result: myvar = 3

so basically what the script must do is this.

I will be calling the function and give it an argument. I want to declare that argument INSIDE the function. In other words, I choose the name of the variable when I call the function.

But it doesn't work...

James · Accepted Answer · 2013-10-23 19:19:21Z

1

Any of the following will work:

let $1=3
export $1=3
local $1=3
eval $1=3

Note that some of these alternatives have side effect, more specifically regarding the visibility of the variable.

edited Oct 23, 2013 at 19:19

answered Oct 23, 2013 at 19:11

James

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3 Comments

Matt Over a year ago

This worked !!!!! But is it possible to know why it's working? What does export do exactly? How can it be used except in these situations?

James Over a year ago

Well, put simply, this is due to the parsing rules of the shell. The parser either try to match the pattern "<variable_name>=<value>" or a command expression. Variable substitution are not acceptable in production rule "<variable_name>", so "$var=12 34 45", after variable substitution, will be interpreted as the name of a program to be executed (something like "bar=12", with argument "34" and "45").

James Over a year ago

As for the "export" buit-in command, in both allocate a value to a variable, and mark the variable to be exported, that is, copied to child processes' environment.

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