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Top Forums Shell Programming and Scripting This is driving me nuts error on line 17 Post 302926282 by osdustin on Saturday 22nd of November 2014 03:52:13 PM
Old 11-22-2014
This is driving me nuts error on line 17

There is an error in this script on line 17 bee at it for 12 hour trying to find the problem, just lost. please help a newbie.
Code:
line 17: [: expr: integer expression expected 1

Code:
 #!/bin/sh
  2 ## Name of the program is test_script_b.sh
  3 
  4 ##request information from the user
  5 echo enter 3 numbers
  6 read a b c
  7 
  8 ##check for proper range of numbers
  9 if [ $a -gt 10 -o $b -gt 10 -o $c -gt 5 ]; then
 10 echo "numbers out of range"
 11 exit 1
 12 fi
 13 
 14 ## process data
 15 x=($a + $b + $c)
 16 
 17 while [ $x -gt 0 ]
 18 do
 19         echo "(($x / 2 % 2))"
 20         x=(expr $x - 1)
 21 done
 22 
 23 ## end of program

~

Last edited by Scrutinizer; 11-22-2014 at 04:56 PM.. Reason: code tags
 

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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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