foo(a, b, c) function parameters (a, b, c) default definition
Here's my code:
function foo(a, b, c) {
// a, b, c default @param ?
};
or foo(a, b, callback) {} default callback ? thanks.
3 Answers
The usual solution is this :
function foo(a, b, c) {
a = a || 22;
b = b || "ee";
c = c || function(){};
};
If your parameters could have 0 or '' as valid values, you may test
if (typeof(a)=='undefined') a==22;
1 Comment
Denys Séguret
Yes, the accepted answer is wrong in my opinion : if OP doesn't be careful he will have hard to find bugs.
CoffeeScript provides great syntax for that:
test = (a = 1, b = 2) ->
console.log a, b
When compiled to JavaScript:
(function() {
var test;
test = function(a, b) {
if (a == null) {
a = 1;
}
if (b == null) {
b = 2;
}
return console.log(a, b);
};
}).call(this);
Comments
It just give in fo function;
function foo(a, b, callback) {
// @param defaults
(a || (a = "default a"));
(b || (b = "default b"));
// if callback not function is function() {} --> callback();
(callback || (callback = function() { return ""}));
console.log(a, b, callback());
};
foo(); // default a default b undefined
foo("a--", "b--", function() { return "callback"}); // a-- b-- callback
or
function foo(a, b, callback) {
// @param defaults
var a = a || "default a";
var b = b || "default b";
// if callback not function is function() {} --> callback();
var callback = callback || function() {};
console.log(a, b, callback());
};
