When double quote is considered when not?
The following scripts were run for /bin/bash, version 4.0.33, and then comes their outputs. In the second example seems to have a warning: binary operator expected. Why the error is generated? and why there is no error for the first example? -- var1=word1 word2 echo $var1 if (test -z \$var1\) then echo first else echo second fi word1 word2 second -- var1=word1 word2 echo $var1 if (test -z $var1) then echo first else echo second fi word1 word2 ./ss1: line 3: test: word1: binary operator expected second --
Re: When double quote is considered when not?
On Wednesday 30 Mar 2011 11:13:58 ali hagigat wrote: The following scripts were run for /bin/bash, version 4.0.33, and then comes their outputs. In the second example seems to have a warning: binary operator expected. Why the error is generated? and why there is no error for the first example? -- var1=word1 word2 echo $var1 if (test -z \$var1\) then echo first else echo second fi word1 word2 second -- var1=word1 word2 echo $var1 if (test -z $var1) then echo first else echo second fi word1 word2 ./ss1: line 3: test: word1: binary operator expected second -- Use set -x at the beginning of the script and rerun it, and you'll immediately see why. -- D.
Re: When double quote is considered when not?
On 30.03.2011 12:13, ali hagigat wrote: The following scripts were run for /bin/bash, version 4.0.33, and then comes their outputs. In the second example seems to have a warning: binary operator expected. Why the error is generated? and why there is no error for the first example? -- var1=word1 word2 echo $var1 if (test -z \$var1\) then why run the test in a subshell? is there any reason for the second quote pair? echo first else echo second fi word1 word2 second -- var1=word1 word2 echo $var1 if (test -z $var1) then expands to nothing, as nothing in between opening and closing quote. $var1 (now unqoted) expands to 'word1 word2', so the test command sees 'word2' while it expects another operator. echo first else echo second fi word1 word2 ./ss1: line 3: test: word1: binary operator expected second --
Re: When double quote is considered when not?
On Wed, Mar 30, 2011 at 02:43:58PM +0430, ali hagigat wrote: -- var1=word1 word2 echo $var1 if (test -z \$var1\) then echo first else echo second fi Problems in this script: * Unquoted $var1 on line 2. This means that instead of passing the actual content of the variable to echo, bash is going to split it into words, perform globbing (filename expansion) on each of those words, and then pass the resulting lists of words to echo. * Unnecessary subshell invocation (parentheses) on line 3. There's no reason to create a new process there -- all it does is slow down the script. * You are testing the length of a string that will NEVER be zero-length. The string you are testing has two literal quotation marks in it, so it will ALWAYS be at least 2 characters long. The test -z will therefore never be true. -- var1=word1 word2 echo $var1 if (test -z $var1) then echo first else echo second fi Problems in this script: * Unquoted $var1 on line 2, same as before. * Unquoted $var1 on line 3. This is even worse than the case on line 2, because when you pass multiple arguments to the test command following a -z switch, you will generate an error condition. * Unnecessary subshell, just as before. * Unnecessary use of double double quotes on line 3. All they are doing is increasing the length of the code. They have no effect. $var1 and $var1 and $var1 are all the same thing. Adding zero-length strings to a string does not change the string, no matter how many of them you add. Here is what you SHOULD have done: var1=what ever echo $var1 if test -z $var1; then echo empty else echo not empty fi
Re: When double quote is considered when not?
On 30.03.2011 14:09, Mart Frauenlob wrote: [...] so the test command sees 'word2' while it expects another operator. sorry, i meant to write 'word1'
