I'm quite confused When I try to ouput some patterns with * in shell.The code is:
#!/bin/bash
for i in {1..10}
do
tmpstr=""
for ((c=1;c<=i;c++))
do
tmpstr=$tmpstr'*'
done
echo $tmpstr #add some string after tmpstr will work
done
The output shows me the result of ls command in each line which is unexpected. And the code will works fine if I add any string after echo $tmpstr.For example,echo $tmpstr" " .So how to understand this?
Your script is generating the following for tmpStr
*
**
***
etc.
which results in the following echo statements
echo *
echo **
echo ***
etc.
The shell interprets the * as a wildcard and expands it by listing all the files in the current directory.
Note that if you put quotes around the shell variable:
echo "$tmpstr"
The shell does not expand the wildcard characters and the output is
*
**
***
****
*****
******
*******
********
*********
**********
echo $tmpstr" " inside the first loop not the inner one.
echo * and then echo *" ". In the first case the shell treats * as a wildcard and expands it. In the second case the contiguous bits * and " " are concatenated into a single word * , and the * within the word is not expanded
echo * the final space is not treated as part of the "word" *; with echo *" " the double-quotes force the shell to treat it as a normal part of the word. If you had any files with names ending in space, echo *" " would list them, but you don't so the wildcard doesn't get expanded. Try this: touch 'foo '; echo *" " (and then try your script with the echo $tmpstr" " again).