Java is always pass-by-value. Unfortunately, when we deal with objects we are really dealing with object-handles called references which are passed-by-value as well. This terminology and semantics easily confuse many beginners.
It goes like this:
public static void main(String[] args) {
Dog aDog = new Dog("Max");
Dog oldDog = aDog;
// we pass the object to foo
foo(aDog);
// aDog variable is still pointing to the "Max" dog when foo(...) returns
aDog.getName().equals("Max"); // true
aDog.getName().equals("Fifi"); // false
aDog == oldDog; // true
}
public static void foo(Dog d) {
d.getName().equals("Max"); // true
// change d inside of foo() to point to a new Dog instance "Fifi"
d = new Dog("Fifi");
d.getName().equals("Fifi"); // true
}
In the example above aDog.getName()
will still return "Max"
. The value aDog
within main
is not changed in the function foo
with the Dog
"Fifi"
as the object reference is passed by value. If it were passed by reference, then the aDog.getName()
in main
would return "Fifi"
after the call to foo
.
Likewise:
public static void main(String[] args) {
Dog aDog = new Dog("Max");
Dog oldDog = aDog;
foo(aDog);
// when foo(...) returns, the name of the dog has been changed to "Fifi"
aDog.getName().equals("Fifi"); // true
// but it is still the same dog:
aDog == oldDog; // true
}
public static void foo(Dog d) {
d.getName().equals("Max"); // true
// this changes the name of d to be "Fifi"
d.setName("Fifi");
}
In the above example, Fifi
is the dog's name after call to foo(aDog)
because the object's name was set inside of foo(...)
. Any operations that foo
performs on d
are such that, for all practical purposes, they are performed on aDog
, but it is not possible to change the value of the variable aDog
itself.
For more information on pass by reference and pass by value, consult the following SO answer: https://stackoverflow.com/a/430958/6005228. This explains more thoroughly the semantics and history behind the two and also explains why Java and many other modern languages appear to do both in certain cases.
To check if a directory exists in a shell script, you can use the following:
if [ -d "$DIRECTORY" ]; then
# Control will enter here if $DIRECTORY exists.
fi
Or to check if a directory doesn't exist:
if [ ! -d "$DIRECTORY" ]; then
# Control will enter here if $DIRECTORY doesn't exist.
fi
However, as Jon Ericson points out, subsequent commands may not work as intended if you do not take into account that a symbolic link to a directory will also pass this check.
E.g. running this:
ln -s "$ACTUAL_DIR" "$SYMLINK"
if [ -d "$SYMLINK" ]; then
rmdir "$SYMLINK"
fi
Will produce the error message:
rmdir: failed to remove `symlink': Not a directory
So symbolic links may have to be treated differently, if subsequent commands expect directories:
if [ -d "$LINK_OR_DIR" ]; then
if [ -L "$LINK_OR_DIR" ]; then
# It is a symlink!
# Symbolic link specific commands go here.
rm "$LINK_OR_DIR"
else
# It's a directory!
# Directory command goes here.
rmdir "$LINK_OR_DIR"
fi
fi
Take particular note of the double-quotes used to wrap the variables. The reason for this is explained by 8jean in another answer.
If the variables contain spaces or other unusual characters it will probably cause the script to fail.
Best Answer
The Process object streams are misleadingly named. You need to write to the OutputStream of the Process object.
From the doc:
You also have to consume the process output and standard error concurrently, otherwise buffers will fill up and the process will block. See here for more information.