A lambda is just an anonymous function - a function defined with no name. In some languages, such as Scheme, they are equivalent to named functions. In fact, the function definition is re-written as binding a lambda to a variable internally. In other languages, like Python, there are some (rather needless) distinctions between them, but they behave the same way otherwise.
A closure is any function which closes over the environment in which it was defined. This means that it can access variables not in its parameter list. Examples:
def func(): return h
def anotherfunc(h):
return func()
This will cause an error, because func
does not close over the environment in anotherfunc
- h
is undefined. func
only closes over the global environment. This will work:
def anotherfunc(h):
def func(): return h
return func()
Because here, func
is defined in anotherfunc
, and in python 2.3 and greater (or some number like this) when they almost got closures correct (mutation still doesn't work), this means that it closes over anotherfunc
's environment and can access variables inside of it. In Python 3.1+, mutation works too when using the nonlocal
keyword.
Another important point - func
will continue to close over anotherfunc
's environment even when it's no longer being evaluated in anotherfunc
. This code will also work:
def anotherfunc(h):
def func(): return h
return func
print anotherfunc(10)()
This will print 10.
This, as you notice, has nothing to do with lambdas - they are two different (although related) concepts.
Well, the problem is that the variable i
, within each of your anonymous functions, is bound to the same variable outside of the function.
ES6 solution: let
ECMAScript 6 (ES6) introduces new let
and const
keywords that are scoped differently than var
-based variables. For example, in a loop with a let
-based index, each iteration through the loop will have a new variable i
with loop scope, so your code would work as you expect. There are many resources, but I'd recommend 2ality's block-scoping post as a great source of information.
for (let i = 0; i < 3; i++) {
funcs[i] = function() {
console.log("My value: " + i);
};
}
Beware, though, that IE9-IE11 and Edge prior to Edge 14 support let
but get the above wrong (they don't create a new i
each time, so all the functions above would log 3 like they would if we used var
). Edge 14 finally gets it right.
ES5.1 solution: forEach
With the relatively widespread availability of the Array.prototype.forEach
function (in 2015), it's worth noting that in those situations involving iteration primarily over an array of values, .forEach()
provides a clean, natural way to get a distinct closure for every iteration. That is, assuming you've got some sort of array containing values (DOM references, objects, whatever), and the problem arises of setting up callbacks specific to each element, you can do this:
var someArray = [ /* whatever */ ];
// ...
someArray.forEach(function(arrayElement) {
// ... code code code for this one element
someAsynchronousFunction(arrayElement, function() {
arrayElement.doSomething();
});
});
The idea is that each invocation of the callback function used with the .forEach
loop will be its own closure. The parameter passed in to that handler is the array element specific to that particular step of the iteration. If it's used in an asynchronous callback, it won't collide with any of the other callbacks established at other steps of the iteration.
If you happen to be working in jQuery, the $.each()
function gives you a similar capability.
Classic solution: Closures
What you want to do is bind the variable within each function to a separate, unchanging value outside of the function:
var funcs = [];
function createfunc(i) {
return function() {
console.log("My value: " + i);
};
}
for (var i = 0; i < 3; i++) {
funcs[i] = createfunc(i);
}
for (var j = 0; j < 3; j++) {
// and now let's run each one to see
funcs[j]();
}
Since there is no block scope in JavaScript - only function scope - by wrapping the function creation in a new function, you ensure that the value of "i" remains as you intended.
Best Answer
No. Only class instances can be referred to via weak references in Swift, and a function is not a class instance. (And not only must they be class instances, they must be an Optional wrapping a class instance.)
There are some pretty obvious ways around this, or course - the simplest being a wrapper class. But I do not actually recommend that in this situation, because you have not convinced me that weak references to functions are needed here in the first place. Remember, a weak reference to an object to which there is no strong reference will instantly lose the reference and will be pointing at nil. I can't believe that is what you want. I think you're barking up a wrong tree here.