Declaration
A prototype for a function which takes a function parameter looks like the following:
void func ( void (*f)(int) );
This states that the parameter f
will be a pointer to a function which has a void
return type and which takes a single int
parameter. The following function (print
) is an example of a function which could be passed to func
as a parameter because it is the proper type:
void print ( int x ) {
printf("%d\n", x);
}
Function Call
When calling a function with a function parameter, the value passed must be a pointer to a function. Use the function's name (without parentheses) for this:
func(print);
would call func
, passing the print function to it.
Function Body
As with any parameter, func
can now use the parameter's name in the function body to access the value of the parameter. Let's say that func
will apply the function it is passed to the numbers 0-4. Consider, first, what the loop would look like to call print directly:
for ( int ctr = 0 ; ctr < 5 ; ctr++ ) {
print(ctr);
}
Since func
's parameter declaration says that f
is the name for a pointer to the desired function, we recall first that if f
is a pointer then *f
is the thing that f
points to (i.e. the function print
in this case). As a result, just replace every occurrence of print in the loop above with *f
:
void func ( void (*f)(int) ) {
for ( int ctr = 0 ; ctr < 5 ; ctr++ ) {
(*f)(ctr);
}
}
Source
Arguments are passed by assignment. The rationale behind this is twofold:
- the parameter passed in is actually a reference to an object (but the reference is passed by value)
- some data types are mutable, but others aren't
So:
If you pass a mutable object into a method, the method gets a reference to that same object and you can mutate it to your heart's delight, but if you rebind the reference in the method, the outer scope will know nothing about it, and after you're done, the outer reference will still point at the original object.
If you pass an immutable object to a method, you still can't rebind the outer reference, and you can't even mutate the object.
To make it even more clear, let's have some examples.
List - a mutable type
Let's try to modify the list that was passed to a method:
def try_to_change_list_contents(the_list):
print('got', the_list)
the_list.append('four')
print('changed to', the_list)
outer_list = ['one', 'two', 'three']
print('before, outer_list =', outer_list)
try_to_change_list_contents(outer_list)
print('after, outer_list =', outer_list)
Output:
before, outer_list = ['one', 'two', 'three']
got ['one', 'two', 'three']
changed to ['one', 'two', 'three', 'four']
after, outer_list = ['one', 'two', 'three', 'four']
Since the parameter passed in is a reference to outer_list
, not a copy of it, we can use the mutating list methods to change it and have the changes reflected in the outer scope.
Now let's see what happens when we try to change the reference that was passed in as a parameter:
def try_to_change_list_reference(the_list):
print('got', the_list)
the_list = ['and', 'we', 'can', 'not', 'lie']
print('set to', the_list)
outer_list = ['we', 'like', 'proper', 'English']
print('before, outer_list =', outer_list)
try_to_change_list_reference(outer_list)
print('after, outer_list =', outer_list)
Output:
before, outer_list = ['we', 'like', 'proper', 'English']
got ['we', 'like', 'proper', 'English']
set to ['and', 'we', 'can', 'not', 'lie']
after, outer_list = ['we', 'like', 'proper', 'English']
Since the the_list
parameter was passed by value, assigning a new list to it had no effect that the code outside the method could see. The the_list
was a copy of the outer_list
reference, and we had the_list
point to a new list, but there was no way to change where outer_list
pointed.
String - an immutable type
It's immutable, so there's nothing we can do to change the contents of the string
Now, let's try to change the reference
def try_to_change_string_reference(the_string):
print('got', the_string)
the_string = 'In a kingdom by the sea'
print('set to', the_string)
outer_string = 'It was many and many a year ago'
print('before, outer_string =', outer_string)
try_to_change_string_reference(outer_string)
print('after, outer_string =', outer_string)
Output:
before, outer_string = It was many and many a year ago
got It was many and many a year ago
set to In a kingdom by the sea
after, outer_string = It was many and many a year ago
Again, since the the_string
parameter was passed by value, assigning a new string to it had no effect that the code outside the method could see. The the_string
was a copy of the outer_string
reference, and we had the_string
point to a new string, but there was no way to change where outer_string
pointed.
I hope this clears things up a little.
EDIT: It's been noted that this doesn't answer the question that @David originally asked, "Is there something I can do to pass the variable by actual reference?". Let's work on that.
How do we get around this?
As @Andrea's answer shows, you could return the new value. This doesn't change the way things are passed in, but does let you get the information you want back out:
def return_a_whole_new_string(the_string):
new_string = something_to_do_with_the_old_string(the_string)
return new_string
# then you could call it like
my_string = return_a_whole_new_string(my_string)
If you really wanted to avoid using a return value, you could create a class to hold your value and pass it into the function or use an existing class, like a list:
def use_a_wrapper_to_simulate_pass_by_reference(stuff_to_change):
new_string = something_to_do_with_the_old_string(stuff_to_change[0])
stuff_to_change[0] = new_string
# then you could call it like
wrapper = [my_string]
use_a_wrapper_to_simulate_pass_by_reference(wrapper)
do_something_with(wrapper[0])
Although this seems a little cumbersome.
Best Answer
Parameters in calls to functions in PowerShell (all versions) are space-separated, not comma separated. Also, the parentheses are entirely unneccessary and will cause a parse error in PowerShell 2.0 (or later) if
Set-StrictMode
-Version 2
or higher is active. Parenthesised arguments are used in .NET methods only.