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Top Forums Shell Programming and Scripting Problem in arithemetic operation Post 302188386 by era on Wednesday 23rd of April 2008 10:31:08 AM
Old 04-23-2008
expr is picky about its syntax, it wants spaces between everything; in particular, on both sides of parentheses, operators, variables, and, well, everything else. You also need to quote or backslash-escape the parentheses, otherwise you will get a warning from the shell. (Actually that's the warning you are getting at the mo.)

Last edited by era; 04-23-2008 at 11:32 AM.. Reason: Quote parens, too
 

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EXPR(1) 						      General Commands Manual							   EXPR(1)

NAME
expr - evaluate arguments as an expression SYNOPSIS
expr arg ... DESCRIPTION
The arguments are taken as an expression. After evaluation, the result is written on the standard output. Each token of the expression is a separate argument. The operators and keywords are listed below. The list is in order of increasing precedence, with equal precedence operators grouped. expr | expr yields the first expr if it is neither null nor `0', otherwise yields the second expr. expr & expr yields the first expr if neither expr is null or `0', otherwise yields `0'. expr relop expr where relop is one of < <= = != >= >, yields `1' if the indicated comparison is true, `0' if false. The comparison is numeric if both expr are integers, otherwise lexicographic. expr + expr expr - expr addition or subtraction of the arguments. expr * expr expr / expr expr % expr multiplication, division, or remainder of the arguments. expr : expr The matching operator compares the string first argument with the regular expression second argument; regular expression syntax is the same as that of ed(1). The (...) pattern symbols can be used to select a portion of the first argument. Otherwise, the matching operator yields the number of characters matched (`0' on failure). ( expr ) parentheses for grouping. Examples: To add 1 to the Shell variable a: a=`expr $a + 1` To find the filename part (least significant part) of the pathname stored in variable a, which may or may not contain `/': expr $a : '.*/(.*)' '|' $a Note the quoted Shell metacharacters. SEE ALSO
sh(1), test(1) DIAGNOSTICS
Expr returns the following exit codes: 0 if the expression is neither null nor `0', 1 if the expression is null or `0', 2 for invalid expressions. 7th Edition April 29, 1985 EXPR(1)
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