Bash scripting doubt

[bala1486]

Hello,
I have a basic doubt in bash scripting.
I read that "$" symbol is used for expansion. Suppose $var appears in a statement it just substitutes the value of var in that statement.

var=My name is Bala
This is wrong. It needs double quotes as it contains spaces.
right one is
var="My name is Bala"

Suppose
var="My name is Bala"
con=$var
I thought that the second line is wrong and it should be con="$var".
But bash prints My name is Bala if echo $con is typed. How is this possible?
The line should have expanded as con=My name is Bala if just substitution is used. What am i misunderstanding in this?

Also command like var=$(ls | grep data) also works the same as var="$(ls | grep data)" even if the result has more than 1 word.

Please explain me.. Thank you..

Thanks,
Bala

[sixdrift]

I may get the subtlety of this incorrect, but essentially you have to think of the variable assignment in terms of lvalue = rvalue just as you would in any compiled language. The rvalue is a pointer to the stored value that was obtained during the language scan.

In BASH, the rvalue scan of

var=My name is Bala

does not work because the first space character terminates the token scan. The remaining words are then expected to be a new command string.

But the rvalue scan of

var="My name is Bala"

does work because the quoting allows the token scan to continue over the spaces and only complete after the second quote.

Now when BASH encounters

con=$var

it is not re-scanning a textual substitution of what $var is set to, at that time it is assigning the stored value of $var that was already scanned to $con.

Now if you quote a variable on the right hand side, it is subject to quoting rules. I think in the case of

con="$var"

what is happening is that $var is being evaluated, then substituted literally, and then scanned. The result is essentially the same, but the quoting changes the scanning behavior. Chapter 5 of the Advanced Bash Scripting Guide has more on quoting.