javascript，每当我尝试计算参数的平均值时，当它为假时返回真

Question

function passingAverage(n){
    var average = 0, i;
    for (i=0; i < n.length; i+= 1){
        average += n[i];
    }
    if ((average/n.length)<49)
        return false;
    else
        return true;
}

in this code, I compare the numbers given in the console.log and just calculate for average.

console.log('passingAverage(75,42,98) returns: ' + passingAverage(75,42,98));
console.log('passingAverage(34,93,77,89,49) returns: ' + passingAverage(34,93,77,89,49));
console.log('passingAverage(33,61) returns: ' + passingAverage(33,61));

console.log("\n");

for console.log(passing average(75,42,98) and (34,93,77,89,49) indeed does return true which is correct. But, for the console log(33,61), when the 33+61/2 isn't clearly above 49, it returns true automatically for some reason when it should return false. What could be the reason?

Answer 1

That's because you're not passing an array as expected. passingAverage(33,61) pass two parameters.

If you want the function to have the same signature as you define it in the example you provided, then you should call it like this:

passingAverage([33,61]); // pass an array of elements

If you want to call the funtion like passingAverage(33,61) then the function should be defined like this:

function passingAverage(){ // no arguments, we will use the arguments array
    var n = arguments; // arguments is defined for every function it's an array of all the arguments passed in

    var average = 0, i;
    for (i=0; i < n.length; i+= 1){
        average += n[i];
    }
    if ((average/n.length)<49)
        return false;
    else
        return true;
}

// then call the function like this
passingAverage(1, 65, 23, 22); // without using an array

Answer 2

This is a perfect example of why it's best to use debugger rather than console.logs to debug your code. Your function is always returning true because your average is always NaN :

 function passingAverage(n) { var average = 0, i; for (i = 0; i < n.length; i += 1) { average += n[i]; } return average / n.length } console.log('passingAverage(75,42,98) returns: ' + passingAverage(75,42,98)); console.log('passingAverage(34,93,77,89,49) returns: ' + passingAverage(34,93,77,89,49)); console.log('passingAverage(33,61) returns: ' + passingAverage(33,61));

This is because you're passing two distinct float arguments to your method, which is in fact expecting an array of values:

 function passingAverage(n) { console.log("arguments:", n); var average = 0, i; for (i = 0; i < n.length; i += 1) { average += n[i]; } return average / n.length } passingAverage(75,42,98);

Simply pass your values as an array and you'll be good to go.

Answer 3

use arguments to loop over ALL function arguments regardless of length.

function passingAverage() {
    var average = 0, i;
    for (i = 0; i < arguments.length; i++) {
        average += arguments[i];
    }
    if ((average/n.length)<49)
        return false;
    else
        return true;
}