Or, in other words, how to make this work:
function foo(){}
//do something that modifies foo as if it was defined with "function foo(a,b,c){};"
console.log(foo.length);
//output: 3
4 Answers
It is possible, but maybe not very nice:
function lengthDecorator(fun) {
function update(len) {
var args = []; // array of parameter names
for (var i = 0; i < len; ++i) {
args.push('a' + i);
}
var result = new Function('fun',
'return function(' + args.join(',') + ') {' +
'var args = Array.prototype.slice.call(arguments);' +
'return fun.apply(this, args);' + // call supplied function
'}'
); // create a function that will return a function
result = result(fun); // make the fun param known to the inner function
result.update = update;
return result;
}
return update(fun.length);
}
Example usage:
var foo = lengthDecorator(function(a,b) {
return a+b;
});
print('foo.length: ' + foo.length);
print('foo(2, 3): ' + foo(2, 3));
print('');
foo = foo.update(42);
print('foo.length: ' + foo.length);
print('foo(2, 3): ' + foo(2, 3));
Output:
foo.length: 2 foo(2, 3): 5 foo.length: 42 foo(2, 3): 5
(Live demo: Ideone.com, jsFiddle)
lengthDecorator wraps the supplied function with a function that takes the same amount of parameters as the supplied function. The parameter count can be changed with update.
C.f.
new Function(...): Dynamically create a new function.fun.apply(...): "Calls a function with a given this value and arguments provided as an array."
Comments
function foo() {}
alert(foo.length); // 0
foo = function (a, b, c) {}
alert(foo.length); // 3
1 Comment
Šime Vidas
The question implies that the function body should not be replaced.
I'm not sure what you're actually trying to do, but you can store the old foo in var and then redefine foo.
function foo() {...}
var oldfoo = foo;
foo = function (a, b, c) {
oldfoo();
}
But what's the point?
2 Comments
MaiaVictor
This won't work (because I don't know the length before running the code). I'm trying to write a decorator but I'm having trouble with the fact the function returned by it is reporting a wrong value for function.length
bhamlin
@Dokkat perhaps you could share more of your code? It's hard to know what exactly you're doing or why the
foo.length value would matter at all.The length property of a function object is non-writable and non-configurable, so there is no way to change its value.
You could define a new function which invokes the original function internally...
a,b, andc. If not, what's the point of defining them? If yes, why didn't it specify them in the first place?