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RazerJs
Asked: 2020-10-08 03:09:04 +0800 CST 2020-10-08 03:09:04 +0800 CST

Why does my object change?

  • 772

I would like to know why the following happens... Making a function that takes the value of the middle of three numbers I tried with the first idea that came to my mind is the function that is here below...

    function valorMedio (inputArray) {
        console.log(`arreglo original: [${inputArray}]`);
        let mid = inputArray.sort((a,b) => a-b)[1]
        console.log(`arreglo final: [${inputArray}]`);
        return inputArray.indexOf(mid)
    };

    console.log(valorMedio([15, 22, -7]));
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is simple; sort the array take the second number of the sorted array and assign it to a variable and then look up the index of this number in the original array. The problem is that the mutated array returns me index 1 when it should be index 0 , regardless of whether what I'm doing is taking a value from my original array, it's sorted... Which if I do eg: mid =inputArray.sort((a,b) => a-b).join('')and all the complications don't happen later .. Normally if I want to change an array that I receive by parameter I do inputArray = inputArray.sort((a,b) => a-b) array = array.modificaciones().

function valorMedio (inputArray) {
    console.log(`arreglo original: [${inputArray}]`);
    let mid = [...inputArray].sort((a,b)=>a-b)[1]
    
    console.log(`arreglo final: [${inputArray}]`);
    return inputArray.indexOf(mid)
};

console.log(valorMedio([15, 22, -7]));
console.log(valorMedio([1, -22, 75]));
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I thought it was not strange I was assigning a reference... and I solved it by copying the array from this function above but the strange thing was that if I do:

function valorMedio2(a) {
    return a.indexOf(a.sort((a, b) => a - b )[1])
}


function valorMedio3(a) {
    return a.indexOf(a.concat().sort((a, b) => a - b )[1])
}

console.log('Incorrecto '+valorMedio2([15, 22, -7]));
console.log('Correcto '+valorMedio3([15, 22, -7]));
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WITH IT concat()I SOLVE THE PROBLEM... What happens in Javascript that mutates the array???

javascript

1 Answers

  1. Best Answer

    The method sortmutates the array and returns it. So it doesn't matter if you try to save the sort result to a new variable or not, since you'll still be working with the original array:

    const a = [3,5,2,1,4];

const b = a.sort((a,b)=> a-b);

console.log(a.toString(), a === b); //a y b son la misma instancia

    On the other hand, the method concatcreates a new array, so in each function you are asking it to search by index in two different arrays. So you can see the difference I have rewritten your functions making each step more explicit:

    function valorMedio2(a) {
    return a.indexOf(a.sort((a, b) => a - b )[1])
}


function valorMedio3(a) {
    return a.indexOf(a.concat().sort((a, b) => a - b )[1])
}

function valorMedio2Expandida(a) {
    a.sort((a, b) => a - b );
    const elementoMedio = a[1];
    return a.indexOf(elementoMedio); //tiene que ser 1!
}


function valorMedio3Expandida(a) {
    const copiaDeA= a.concat();
    copiaDeA.sort((a, b) => a - b );
    const elementoMedio = copiaDeA[1];
    return a.indexOf(elementoMedio); //devuelve la posición original
}

console.log(valorMedio3Expandida([1,3,2])); //el dos está en la posición 2

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