Noah Noah - 1 year ago 94
Swift Question

Decorators in Swift

I'm new to Swift, and I'd like to know if the language has some equivalent to Python's decorator pattern.
For example:

import functools

def announce(func):
"""Print a function's arguments and return value as it's called."""
def announced_func(*args, **kwargs):
rv = func(*args, **kwargs)
print('In: {0}, {1}'.format(args, kwargs))
print('Out: {}'.format(rv))
return rv
return announced_func

@announce # add = announce(add)
def add(a, b):
return a + b

add(2, 5)
# In: (2, 5), {}
# Out: 7
# 7

Perhaps I just haven't found it yet, but Swift doesn't seem to have a way to forward arbitrary arguments to functions or to preserve a wrapped function's information (as functools.wraps does).

Is there an equivalent, or is the pattern not meant to be used in Swift?

Answer Source

You can use this:

func decorate<T, U>(_ function: @escaping (T) -> U, decoration: @escaping (T, U) -> U) -> (T) -> U {
    return { args in
        decoration(args, function(args))

let add: (Int, Int) -> Int = decorate(+) { args, rv in
    print("In: \(args)")
    print("Out: \(rv)")
    return rv

add(2, 5) // In: (2, 5)\nOut: 7

Or announce as function instead of closure, allowing reuse:

func announce<T, U>(input args: T, output rv: U) -> U {
    print("In: \(args)")
    print("Out: \(rv)")
    return rv

let add: (Int, Int) -> Int = decorate(+, decoration: announce)

add(2, 5) // In: (2, 5)\nOut: 7

let length = decorate({(str: String) in str.characters.count }, decoration: announce)
length("Hello world!") // In: Hello world!\nOut: 12
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