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what do u mean by $#,$* in unix programming?
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Answer Posted / satyabrata
$# it contains the no of positional parameters
$* it contains all positional parameters as a single string
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Shifting positional parameter in Linux fedora core ? Hi I have written following shell script for display value of positional. But in 11th and 12 field will display without shifting command. May I know is it advance of Linux of programming code error? #!/usr/bin/bash echo "Bellow is the out of ps command" echo "`ps`" echo "The passing Parameter i.e output of \$1,2... value is:==> $11 " echo "Total number of passed argument \$# is:==> $#" echo "Passed argument names (\$*) are:==>$*" echo "This script PID(\$$) is :=>$$" echo "The name of executing script(\$0) is :==>$0" echo "The Parent ID of this script(\$PPID) is:==>$PPID" And my input to this script is ./scriptname arg1 arg2 arg3 arg4 arg5 arg6 arg7 arg8 arg9 arg10 arg11 arg12 Output Is Bellow is the out of ps command PID TTY TIME CMD 2892 pts/0 00:00:00 bash 3172 pts/0 00:00:00 positional_para 3173 pts/0 00:00:00 ps The passing Parameter i.e output of $1,2... value is:==> arg11 Total number of passed argument $# is:==> 13 Passed argument names ($*) are:==>arg1 arg2 arg3 arg4 arg5 arg6 arg7 arg8 arg9 arg10 arg11 arg12 arg13 This script PID($$) is :=>3172 The name of executing script($0) is :==>./positional_parameter The Parent ID of this script($PPID) is:==>2892 After the ps output please see the next line I.e $11 value comes out without shifting the parameter. How is it give me my friends
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