Bash pass variable as argument with quotes
<p>Assume <code>./program</code> is a program that just prints out the parameters;</p>

<pre><code>$ ./program "Hello there"
Hello there
</code></pre>

<p>How can I properly pass arguments with quotes in from a variable? I am trying to do this;</p>

<pre><code>$ args='"Hello there"'  
$ echo ${args}  
"Hello there"  
$ ./program ${args}  
Hello there # This is 1 argument
</code></pre>

<p>but instead, when I go through a variable the quotes in <code>args</code> seem to be ignored so I get;</p>

<pre><code>$ args='"Hello there"'
$ echo ${args}
"Hello there"
$ ./program ${args}
"Hello there" # This is 2 arguments
</code></pre>

<p>Is it possible to have bash treat the quotes as if I entered them myself in the first code block?</p>