我正在创建一个程序,从文本文件中读取一些数据,然后获取该数据并查找数字的最小值,最大值和平均值。出于某种原因,我遇到了许多我以前从未见过的荒谬错误。这是我的代码:
import java.io.File;
import java.util.Scanner;
import java.io.FileWriter;
public class Lab1 {
static int count = 0;
static int[] newData2 = new int[count];
// Method for reading the data and putting it into different arrays
static int[] readData() {
File f = new File("data.txt");
int[] newData = new int[100];
try {
Scanner s = new Scanner(f);
while (s.hasNext()) {
newData[count++] = s.nextInt();
}
for (int i = 0; i < newData2.length; i++) {
newData[i] = newData2[i];
return newData2;
}
} catch (Exception e) {
System.out.println("Could not read the file.");
}
}
static int min(int[] newData2) {
int min = newData2[0];
for (int i = 0; i < newData2.length; i++) {
if (newData2[i] < min) {
min = newData2[i];
}
}
return min;
}
static int max(int[] newData2) {
int max = newData2[0];
for (int i = 0; i < newData2.length; i++) {
if (newData2[i] > max) {
max = newData2[i];
}
}
return max;
}
static double average(int[] newData2) {
double average = 0;
int sum = 0;
for (int i = 0; i < newData2.length; i++) {
sum = newData2[i];
}
average = sum / newData2.length;
return average;
}
/*
* static int stddev(int[] newData2) { int[] avgDif = new
* int[newData2.length]; for(int i = 0; i < newData2.length; i++) {
* avgDif[i] = (int) (average(newData2) - newData2[i]); }
*
* }
*/
void write(String newdata, int min, int max, double average, int stddev) {
try {
File file = new File("stats.txt");
file.createNewFile();
FileWriter writer = new FileWriter("stats.txt");
writer.write("Minimum: " + min + "Maximum: " + max + "Average: " + average);
writer.close();
}catch(Exception e) {
System.out.println("Unable to write to the file.");
}
public static void main(String[] args) {
}
}
}
我的readData
方法出错了,它告诉我:
此方法必须返回int []的结果类型。
我实际上是返回一个int数组,所以我不明白这里的问题是什么。
然后在我的main方法中,它表示void是变量main的无效类型。
答案 0 :(得分:0)
static int[] readData() {
File f = new File("data.txt");
int[] newData = new int[100];
try {
Scanner s = new Scanner(f);
while (s.hasNext()) {
newData[count++] = s.nextInt();
}
for (int i = 0; i < newData2.length; i++) {
newData[i] = newData2[i];
return newData2;
}
} catch (Exception e) {
System.out.println("Could not read the file.");
}
//TODO: return something here if there is some kind of error
}
由于try-catch块,您需要考虑可能发生的每种可能性。当程序成功时返回数组时,您期望返回,但是当存在异常时程序仍然需要返回值,但是您没有提供返回值。
答案 1 :(得分:0)
以下是一些提示:
new Scanner(f);
抛出异常,则不会达到第一个返回,所以你需要一个默认值,如下所示:private int[] readData() {
File f = new File("data.txt");
int count = 0;
int[] newData = new int[100];
try {
Scanner s = new Scanner(f);
while (s.hasNext()) {
newData[count++] = s.nextInt(); // maybe you should handle the case where your input is too large for the array "newData"
}
return Arrays.copyOf(newData, count);
} catch (Exception e) {
System.out.println("Could not read the file.");
}
return null;
}
Arrays.copyOf
(见下文)min
和max
假设数组中有元素(至少有一个),您不应该这样做(或使用if测试):private int min(int[] data) {
int min = Integer.MAX_VALUE; // handy constant :)
for (int i = 0; i < data.length; i++) {
if (data[i] < min) {
min = data[i];
}
}
return min;
}
private int max(int[] data) {
int max = 0;
for (int i = 0; i < data.length; i++) {
if (data[i] > max) {
max = data[i];
}
}
return max;
}
average
方法中,有一些错误:private double average(int[] data) {
int sum = 0;
for (int i = 0; i < data.length; i++) {
sum += data[i]; // here if you want a sum it's += not =
}
return (1.0 * sum) / data.length; // you want a double division, local "average" was useless
}
for (int value : newData) {
// use value
}
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