I want to pass an array parameter to a function in bash, and writing some testing code as：
#!/bin/sh
function foo {
a=$1;
for i in ${a[@]} ; do
echo $i
done
}
names=(jim jerry jeff)
foo ${names[@]}
the above code just show jim, rather than the three j*. so my question is:
#!/bin/bash
function foo {
a=($*)
for i in ${a[@]}
do
echo $i
done
}
names=(jim jerry jeff)
foo ${names[@]}
Your code did not show jim to me, but "names", literally. You have to pass the whole array. And you have to recapture it with a=$($).
The manpage part in bash about Arrays is rather long. I only cite one sentence:
Referencing an array variable without a subscript is equivalent to referencing the array with a subscript of 0.
You're fairly close; the biggest problem was the command a=$1, which assigns only the first parameter ($1) to a, while you want to assign the entire list of parameters ($@), and assign it as an array rather than as a string. Other things I corrected: you should use double-quotes around variables whenever you use them to avoid confusion with special characters (e.g. spaces); and start the script with #!/bin/bash, since arrays are a bash extension, not always available in a brand-X shell.
#!/bin/bash
function foo {
a=("$@")
for i in "${a[@]}" ; do
echo "$i"
done
}
names=(jim jerry jeff "jim bob")
foo "${names[@]}"
For example like this:
my_array[0]="jim"
my_array[1]="jerry"
function foo
{
#get the size of the array
n=${#my_array[*]}
for (( Idx = 0; Idx < $n; ++Idx )); do
echo "${my_array[$Idx]}"
done
}
