Getting number of argument passed to a shell script
Hi Experts,
I have been trying to work on a simple shell script that will just add the two argument passed to it. Here is what i tried :
Code:
#!/bin/bash
welcome(){
echo "Welcome to this Progg. which will accept two parameter"
}
main_logic(){
arg=$#
echo "Number of argument passed is $arg"
first_arg=$1
second_arg=$2
echo "First argument is $first_arg"
echo "Second argument is $second_arg"
}
welcome
main_logic
Below is the output which isn't correct :-
Code:
Welcome to this Progg. which will accept two parameter
Number of argument passed is 0
First argument is
Second argument is
Yes, and as you have already received the solution i wil concentrate on explaining the reason:
When you invoke a process with some arguments these arguments become part of the "process environment". Inside a script you can then use special variables ("$@", "$*", "$1", "$2", ...) to pull these arguments out of the environment or get ("$#") the number of parameters passed.
But every time you invoke a shell function inside a script this function gets its own environment of sorts and all the special variables "$1", "$2", etc. and "$#" describe what is passed as argument to the function instead of what is passed to the main script. Look at the following sample script:
Code:
#! /bin/bash
subfunc()
{
echo "inside subfunc"
echo "Number of arguments passed: $#"
echo "argument 1: $1"
echo "argument 2: $2"
echo "argument 3: $3"
}
# main part
echo "inside main"
echo "Number of arguments passed: $#"
echo "argument 1: $1"
echo "argument 2: $2"
echo "argument 3: $3"
subfunc a b c
subfunc X Y
And its output will be:
Code:
# ./sample.sh foo bar baz blah
inside main
Number of arguments passed: 4
argument 1: foo
argument 2: bar
argument 3: baz
inside subfunc
Number of arguments passed: 3
argument 1: a
argument 2: b
argument 3: c
inside subfunc
Number of arguments passed: 2
argument 1: X
argument 2: Y
argument 3:
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