How to create an object from an Array of key-value pairs?

You can use the reduce function

x = [[1,2],[3,4],[5,6]];
o = x.reduce(function(prev,curr){prev[curr[0]]=curr[1];return prev;},{})

o is now the equivalent of {1:2, 3:4, 5:6}

If your input array is sparse, you'll want to add a if(curr!=undefined) test on the assignment, but make sure you still return "prev".

If your tuples are something more complex than simple [key,value], you can simply change the way "prev" is assigned to. Eg: prev["key-"+curr[0].toString()]=curr[1]*2;

Object.fromEntries does the job. It was added to the language with EcmaScript2019.

If you don't have support for that function, you could define it yourself with the following ES2015 code:

Object.fromEntries = arr => Object.assign({}, ...Array.from(arr, ([k, v]) => ({[k]: v}) ));

A nice thing is that this method does the opposite of Object.entries (ES2017), so now you can go back and forth between the object and array representation:

const arr = [['name', 'Bob'], ['age', 42], ['breakfast', 'eggs']];
const obj = Object.fromEntries(arr);
console.log(obj);
// ... and back:
const arr2 = Object.entries(obj);
console.log(arr2); // copy of the original array (omitting duplicate keys)

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ES6 Map

There is an alternative to plain objects for key/value pairs: Map.

Its constructor accepts the array-of-pairs format:

// Demo:
const arr = [['name', 'Bob'], ['age', 42], ['breakfast', 'eggs']];

const mp = new Map(arr);

// Get one particular value:
console.log(mp.get('age'));
// Get all the keys:
console.log(...mp.keys());
// Get all the values:
console.log(...mp.values());
// Get all the key/value pairs:
console.log(...mp.entries());

If you really need a plain object, then this is not useful, but a Map might present a viable alternative.

At the time of writing (2013) JavaScript objects / dictionaries / associative arrays don't have such a constructor natively.

As you said yourself, you can of course build your own function using for instance a functional approach using the reduce function as explained in one of the other answers. A classic for or newer forEach loop would also work, of course. But there isn't anything built-in.

Edit: It's 2019 and now we have Object.fromEntries, which will give you what you need.