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Full Discussion: How to remove empty line.?
Top Forums Shell Programming and Scripting How to remove empty line.? Post 302975927 by Scrutinizer on Tuesday 21st of June 2016 02:46:32 PM
Old 06-21-2016
You are welcome. If you try:
Code:
( set -x; [ -n $p ] && [ -n "$p" ])
+ '[' -n ']'
+ '[' -n '' ']'

You'll notice that the first one evaluates to -n and the second one to -n ''

In the first case the there is no protection of variable p, which contains an empty field. Since there is not protective quotes around $p, the result is field split and since $p is empty this results in 0 fields. So wat remains "-n" which is a test if the string "-n" is non empty
([ "$somestring" ] if the same as [ -n "$somestring"], I prefer the latter for clarity, so in fact [ -n ] is equivalent to [ -n -n ])


In the second case the result of "$p" is not field split and so the -n is interpreted as an operator with one field as parameter, which contains an empty variable, and this evaluates to false .

So that is yet another reason to put quotes around variable expansions.

I hope this clarifies things...

Last edited by Scrutinizer; 06-21-2016 at 03:55 PM..
This User Gave Thanks to Scrutinizer For This Post:
 

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EXPR(1) 						    BSD General Commands Manual 						   EXPR(1)

NAME
expr -- evaluate expression SYNOPSIS
expr expression DESCRIPTION
The expr utility evaluates expression and writes the result on standard output. All operators are separate arguments to the expr utility. Characters special to the command interpreter must be escaped. Operators are listed below in order of increasing precedence. Operators with equal precedence are grouped within { } symbols. expr1 | expr2 Returns the evaluation of expr1 if it is neither an empty string nor zero; otherwise, returns the evaluation of expr2. expr1 & expr2 Returns the evaluation of expr1 if neither expression evaluates to an empty string or zero; otherwise, returns zero. expr1 {=, >, >=, <, <=, !=} expr2 Returns the results of integer comparison if both arguments are integers; otherwise, returns the results of string comparison using the locale-specific collation sequence. The result of each comparison is 1 if the specified relation is true, or 0 if the relation is false. expr1 {+, -} expr2 Returns the results of addition or subtraction of integer-valued arguments. expr1 {*, /, %} expr2 Returns the results of multiplication, integer division, or remainder of integer-valued arguments. expr1 : expr2 The ``:'' operator matches expr1 against expr2, which must be a regular expression. The regular expression is anchored to the begin- ning of the string with an implicit ``^''. expr expects "basic" regular expressions, see re_format(7) for more information on regu- lar expressions. If the match succeeds and the pattern contains at least one regular expression subexpression ``(...)'', the string corresponding to ``1'' is returned; otherwise the matching operator returns the number of characters matched. If the match fails and the pattern contains a regular expression subexpression the null string is returned; otherwise 0. Parentheses are used for grouping in the usual manner. EXAMPLES
1. The following example adds one to the variable a. a=`expr $a + 1` 2. The following example returns the filename portion of a pathname stored in variable a. The // characters act to eliminate ambiguity with the division operator. expr //$a : '.*/(.*)' 3. The following example returns the number of characters in variable a. expr $a : '.*' DIAGNOSTICS
The expr utility exits with one of the following values: 0 the expression is neither an empty string nor 0. 1 the expression is an empty string or 0. 2 the expression is invalid. STANDARDS
The expr utility conforms to IEEE Std 1003.2 (``POSIX.2''). BSD
July 3, 1993 BSD
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