Javascript-如何通过将数组的所有元素相乘来返回正确的值？

Question

What I'm trying to do here it's multiply all the elements of the array A. With these values: [1,2,0,-5]it should returns 0... but it returns 1. What I'm doing wrong? Here's the code:

function solution(A){
    let multi = 1;
    for(i = 1; i < A.length; i++){
        multi *= A[i]
    } 

    if(multi = 30){
        return 1
    } else if (multi = -30){
        return -1
    } else if (multi = 0){
        return 0
    } else{
        console.log("hey hey");
    }
}

solution(A = [1,2,0,-5])

Answer 1

Your loop is starting at 1 - JavaScript arrays (and arrays in many other languages) are 0-indexed, so start at 0 . Your if conditions are also wrong - use the comparison operator == not the assignment operator = .

function solution(A){
    let multi = 1;
    for(i = 0; i < A.length; i++){
        multi *= A[i]
    } 

    if(multi == 30){
        return 1
    } else if (multi == -30){
        return -1
    } else if (multi == 0){
        return 0
    } else{
        console.log("hey hey");
    }
}

Answer 2

Javascript arrays start at 0 not 1. Anyway since that is causing trouble you should use the "reduce" function to multiple the elements of an array as this avoids the need worrying about the indices.

let mult = A.reduce((a,b)=>a*b);

With the if there is an assignment not a comparison. if(multi=30) then tests whether multi is non-zero, which it is because it is 30. One way to avoid this problem is putting the constant on the left:-

  if(30 === multi){
        return 1
    } else if (-30 ===multi){
        return -1
    } else if (0===multi){
        return 0
    } else{
        console.log("hey hey");
    }

Also passing values to functions is by position. No need to name the parameter, it is incorrect to do that.

solution([1,2,0,-5]);

Answer 3

 function solution(A){ const mult = A.reduce((a, b) => a*b); return mult == 30 ? 1 : mult == -30 ? -1 : 0; } console.log(solution(A = [1,2,0,-5]));