what is the command To print script arguments
Answers were Sorted based on User's Feedback
Answer / shabab
$* and $@ are the right answers
$# - prints out the number of arguments passed
Consider the below code
########################
for i in "$*"
do
print $i
done
for i in "$@"
do
print $i
done
########################
and you call the script by saying
#samp.sh hai welcome to "Unix Forum"
hai welcome to Unix Forum
hai
welcome
to
Unix Forum
The first line is the output by printing out $*
the next four lines are with the help of $@.
So
$* will combine all arguments to a single string
$@ will have each arguments as a seperate string
Is This Answer Correct ? | 4 Yes | 0 No |
Answer / bc
$@ seems to be another option. Difference between $* and $@
is that $@ recognizes multiple strings given in quotes as a
single word.
Is This Answer Correct ? | 3 Yes | 0 No |
Answer / alf55
There is a difference between using $@ and using "$@". The
first is the same as using $*, while the latter is what was
being described ad $@. It only handles the arguments
correctly when used as "$@". However, you will not see where
the arguments are changing in its simple usage in a print.
echo "arguments are:"; for arg in "$@"; do echo "
${arg}"; done
Will show each argument on a new line indented by four spaces.
Here is an example:
[code]
bash$ function show_simple_args
> {
> echo "There are $# arguments passed, can you find
them correctly?"
> echo "using \$*:"
> echo $*
> echo "using \$@:"
> echo $@
> echo "using \"\$@\":"
> echo "$@"
> echo "using for loop with \$*:"
> echo "arguments are:"; for arg in $*; do echo "
${arg}"; done
> echo "using for loop with \$@:"
> echo "arguments are:"; for arg in $@; do echo "
${arg}"; done
> echo "using for loop with \"\$@\":"
> echo "arguments are:"; for arg in "$@"; do echo "
${arg}"; done
> }
bash$
bash$ show_simple_args "arg 1" "arg 2" "arg 3" "arg 4"
There are 4 arguments passed, can you find them correctly?
using $*:
arg 1 arg 2 arg 3 arg 4
using $@:
arg 1 arg 2 arg 3 arg 4
using "$@":
arg 1 arg 2 arg 3 arg 4
using for loop with $*:
arguments are:
arg
1
arg
2
arg
3
arg
4
using for loop with $@:
arguments are:
arg
1
arg
2
arg
3
arg
4
using for loop with "$@":
arguments are:
arg 1
arg 2
arg 3
arg 4
bash$
[/code]
Is This Answer Correct ? | 0 Yes | 1 No |
How do I find previous commands in linux?
what is initrd image?
What could be the problem when a command that was issued gave a different result from the last time it was used?
What does pwd mean in linux?
Explain command grouping in linux?
What is umask in linux?
What are the examples of simple command?
You locate a command in the /bin directory but do not know what it does. What command can you use to determine its purpose
How do I check disk space in bash?
What is makefile in unix?
What are whois tools?
What file type is a makefile?