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Create a function that accepts a single parameter

Written byPhuoc Nguyen
15 Mar, 2021
Last updated
08 Sep, 2022
Quite often, we use the `map` function to transform each item of an array into a new one. However, it's common to see an issue if we don't pass the parameter to the mapper function.
For example, the following code converts each item of array into a number:
['1', '2', '3', '4', '5'].map((v) => parseInt(v));

// [1, 2, 3, 4, 5]
However, the result isn't correct if we shorten it as below:
['1', '2', '3', '4', '5'].map(parseInt);

// [1, NaN, NaN, NaN, NaN]
The issue is caused by the fact that the mapper function accepts three parameters which are the array item, index, and the array. Calling `.map(parseInt)` means that we pass the item index to `parseInt` as the second parameter. As a result, we will see `NaN`.
This leads to a requirement of building a function that accepts only the first parameter, and ignore the remaining parameters.
const unary = (fn) => (params) => fn(params);
The `unary` function creates a wrapper of a function, and ignores all parameters except the first one. With that function in our hand, we can pass the mapper to the `map` function like this:
['1', '2', '3', '4', '5'].map(unary(parseInt));

// [1, 2, 3, 4, 5]

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Phước Nguyễn