This question already has answers here :
In JS, the array name only stores the reference to the original array. And sort mutates (changes) the array. You are not making a copy of the array with
const orderBirth = inventors.sort((a, b) => a.year > b.year ? 1 : -1);
Rather, you are storing the reference to the array sorted by year.
When you sort again with different function, the main array changes. And the orderBirth is still pointing to it. So that changes as well.
edit : As Scorpioo590 has said, .slice() method will duplicate the array.
const orderBirth = inventors.slice().sort((a, b) => a.year > b.year ? 1 : -1);
So what you assumed the '=' operator did is correctly achieved in this way.
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Sort
let arr = [1,2,3,5,6,64,76,52]
arr.sort((a,b)=> a-b)
console.log(arr)
The 10 Most Common Mistakes JavaScript Developers Make, At first blush, JavaScript may seem quite simple. As a result, the anonymous function being passed to setTimeout() is being defined in the If you run the above code and monitor memory usage, you'll find that you've got a massive Yes, I know this is the right way sort an array of numbers - but for developers who are� Because of this, the sort() method will produce an incorrect result when sorting numbers. You can fix this by providing a "compare function" (See "Parameter Values" below). Note: This method changes the original array.
sort mutates the original array and returns the sorted array. So, they are all pointing to the same array in memory. Both of these will return true:
console.log(inventors === orderBirth)
console.log(oldest === orderBirth)
const inventors=[{first:'Albert',last:'Einstein',year:1879,passed:1955},{first:'Isaac',last:'Newton',year:1643,passed:1727},{first:'Galileo',last:'Galilei',year:1564,passed:1642},{first:'Marie',last:'Curie',year:1867,passed:1934},{first:'Johannes',last:'Kepler',year:1571,passed:1630},{first:'Nicolaus',last:'Copernicus',year:1473,passed:1543},{first:'Max',last:'Planck',year:1858,passed:1947},{first:'Katherine',last:'Blodgett',year:1898,passed:1979},{first:'Ada',last:'Lovelace',year:1815,passed:1852},{first:'Sarah E.',last:'Goode',year:1855,passed:1905},{first:'Lise',last:'Meitner',year:1878,passed:1968},{first:'Hanna',last:'Hammarström',year:1829,passed:1909}]
const orderBirth = inventors.sort((a, b) => a.year > b.year ? 1 : -1);
const oldest = inventors.sort((a, b) => {
const lastInventor = a.passed - a.year;
const nextInventor = b.passed - b.year;
return lastInventor > nextInventor ? -1 : 1;
});
console.log(inventors === orderBirth)
console.log(oldest === orderBirth)
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As you mentioned the sort functions alters the original array.
With
const orderBirth = inventors.sort(...) you get a reference to the array. Meaning both variables actually point to the same place in memory. Thus when changing the original array you also change the new variable.
You would need to create a copy of the original array to preserve the result of the sorting.
const orderBirth = inventors.sort(...).slice();
should do the trick.
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Comments please have alook here, too, regarding only two values as return value for sorting: stackoverflow.com/q/24080785 I don't think this is a duplicate question, they deal with the same issue but I understood that the sort function mutates the original array (and mentioned it my question). Where I became unstuck was in terms of the variable pointing to the same in-memory array and also the fact that I was never actually re-running the function. Maybe add that this happens even when assigning to a variable. Thanks for the comment but why cant we continue to mutate the array with a different sort? So initially I sort by birth and the array is mutated. Then I sort by age and the array is mutated. Why would it stop there and not allow me to re-mutate by running the sort by birth function again? I think Jack may have given me the clue - I'm not re-running the function, I'm just re-logging the variable! Thanks for your help both.
