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Meg Dooley
7,321 PointsUnderstanding the //array.map() method in JavaScript
const animals = ['Hen', 'elephant', 'llama', 'leopard', 'ostrich', 'Whale', 'octopus', 'rabbit', 'lion', 'dog'];
const secretMessage = animals.map(animals => animals[0]);
console.log(secretMessage.join(''));
Hi all,
Hoping you can help me to understand the above code which prints out "Helloworld" to the console. It works fine its just that I can't figure out why the parameter "animals" has to be the same as the array name. (to clarify, if I change the name to something else the code doesn't work properly). When I look at other examples of the map method this is not the case.
Please help me understand!
3 Answers

jucliofonseca
959 PointsHello Meg!
Actually, the name of parameter doesn't need to be the same of the variable. Please, try the below code:
const animals = ['Hen', 'elephant', 'llama', 'leopard', 'ostrich', 'Whale', 'octopus', 'rabbit', 'lion', 'dog'];
// In the next line it will apply the function to each of elements of animals array and return the first character
const secretMessage = animals.map(a => a[0]);
console.log(secretMessage.join(''));
// Similar Approach:
/*
const secretMessage1 = animals.map(function (a) {return a[0];});
console.log(secretMessage1.join(''));
*/
Best Regards!

Dave StSomeWhere
19,870 PointsThe .map() method is method of the array - so it has to be animals.map()
in your example - the same name as the array.
Could you please show one of the example where it is different.

Meg Dooley
7,321 Pointsvar array1 = [1, 4, 9, 16];
const map1 = array1.map(x => x * 2);
console.log(map1);
Here the parameter name is 'x' - in the example I gave above, the parameter name is the same os the array name and the code doesn't work if I change it.
Meg Dooley
7,321 PointsMeg Dooley
7,321 Pointsah hah! makes sense now thanks!