# Print $1 $2 times
function foo() {
for (( i=0; i<$2; i++ )); do
echo -n $1
done
echo
}
# Print $1 $2x$3 times
function bar() {
for (( i=0; i<$3; i++ )); do
foo $1 $2
done
}
bar $1 $2 $3
The ideal output of foobar.sh @ 3 3 is
@@@
@@@
@@@
but the actual output seems to be just
@@@
Changing the variable in bar() from i to j yields the desired output. But why?
Because variables are "global" in shell-scripts, unless you declare them as local. So if one function changes your variable i, the other function will see these changes and behave accordingly.
So for variables used in functions --especially loop-variables like i, j, x, y-- declareing them as local is a must. See below...
#!/bin/bash
# Print $1 $2 times
function foo() {
local i
for (( i=0; i<"$2"; i++ )); do
echo -n $1
done
echo
}
# Print $1 $2x$3 times
function bar() {
local i
for (( i=0; i<"$3"; i++ )); do
foo "$1" "$2"
done
}
bar "$1" "$2" "$3"
Result:
$ ./foobar.sh a 3 3
aaa
aaa
aaa
$ ./foobar.sh 'a b ' 4 3
a ba ba ba b
a ba ba ba b
a ba ba ba b
