扩展代码我一直在努力寻找补充跟踪器,但我目前的功能并没有返回大于平均值''的精确计数数,也没有返回低于平均值的整数计数。我还在代码中注释了两个问题,因为我不太明白为什么数组被设置为索引[0]。我从评论中学到了很多,并在这里寻找答案。非常感谢这个网站存在!希望通过这个问题有所学习。
function suppArray() {
var nums = new Array(); //create array
var sum = 0; //variable to hold sum of integers in array
var avg = 0; //variable to hold the average
var i;
var count = 0;
var count2 = 0;
var contents = ''; //variable to hold contents for output
var dataPrompt = prompt("How many numbers do you want to enter?", "");
dataPrompt = parseInt(dataPrompt);
for(i = 0; i <= dataPrompt - 1; i++) { //loop to fill the array with numbers
nums[i] = prompt("Enter a number","");
nums[i] = parseInt(nums[i]);
contents += nums[i] + " "; //variable that will be called to display contents
sum = sum + nums[i];
}
avg = sum / nums.length;
for(i = 0; i < nums.length; i++) { //loop to find the largest number
var biggest = nums[0]; //why does this have to be index 0 and not 'i' ?
if(nums[i] > biggest)
biggest = nums[i]; //largest integer in array
}
for(i = 0; i < nums.length; i++) { //loop to find smallest integer
var smallest = nums[0]; //why does this have to be the index 0 and not 'i' ??
if(nums[i] < smallest)
smallest = nums[i]; //smallest integer in array
}
for(count = 0; count < nums.length; count++) { //count of numbers higher than average
if(nums[i] > avg)
count = nums[i];
}
for(count2 = 0; count2 < nums.length; count2++) { //count of numbers lower than average
if(nums[i] < avg)
count2 = nums[i];
}
}
答案 0 :(得分:2)
您的功能并未返回正确的值,因为您正在分配count
或count2
。如果您在最后count
完成代码并且count2
将等于nums.length
。这是因为您在for
循环中使用它们。同样在循环中,您引用i
,此时(我相信)此时也等于nums.length
。
我想你想要这样的东西:
count = 0;
count2 = 0;
for(i = 0; i < nums.length; i++)
{
if(nums[i] > avg)
{
count++; //Increase the count of numbers above the average
}
else if(nums[i] < avg)
{
count2++; //Increase the count of numbers below the average
}
}
您可能希望对scope和for循环进行一些阅读,因为您似乎对它们感到有些困惑。
修改强>
如果你想要数组中最大和最小的值,你可以这样做:
//Assign the values to the first element by default
var biggest = nums[0];
var smallest = nums[0];
for(var i = 1; i < nums.length; i++)
{
//Set biggest to the larger number, either biggest or the current number
biggest = Math.max(biggest, nums[i]);
//Set smallest to the smaller number, either biggest or the current number
smallest = Math.min(smallest, nums[i]);
}
注意:这假设您在数组中至少有一个值