好吧,这是一个学校项目,我似乎无法弄清楚我做错了什么,我意识到有更好的方法可以做很多事情,但必须以这种方式完成(我会注意到具体是什么)。
我需要编写两个非常简单的程序。
第一个程序创建一个包含 300 个 1 到 54 范围内的随机整数的文件,然后打印到文本文件中。
第二个程序读取该文本文件并将内容放入数组中。然后操作该数组以输出一些内容:数字的平均值以及数字在特定范围内出现的次数。我还尝试立即输出整个数组。不幸的是,所有部分基本上都不起作用,或者我不理解某些事情,但我很确定这都是哈哈。
方案1如下
import java.util.*;
import java.io.*;
public class OutStream{
public static void main(String[] args){
int Counter = 0,
Value;
Random Gen = new Random();
System.out.println("Generating Numbers and Text File Now");
try{
File out = new File("F:\\Text.txt");
PrintWriter P = new PrintWriter(out);
while(Counter <= 300){
Value = Gen.nextInt(54)+1;
P.println(Value);
++Counter;
System.out.println("Value: " + Value);
}
P.close();
}catch(Exception E){
E.printStackTrace();
System.out.println(E.getMessage());
}
System.out.println("The Program has ended");
}
}
第二个程序是一切崩溃的地方。
import java.util.*;
import java.io.*;
public class InStream {
public static void main(String[] args) {
int [] NumArray = new int[301];
int PosNum = 0,
Total = 0,
Average = 0,
Range1 = 0,
Range2 = 0,
Range3 = 0,
Range4 = 0,
Range5 = 0,
Range6 = 0;
try{
File Fout = new File("F:\\Text.txt");
PosNum = 0;
Scanner Fin = new Scanner(Fout);
while(Fin.hasNextInt()){
NumArray[++PosNum] = Fin.nextInt();
}
Fin.close();
}catch(Exception E){
E.printStackTrace();
System.out.println("***ERROR ***** " + E.getMessage());
}
for(NumArray[PosNum] = 0; NumArray[PosNum] <= 301;){
Total += NumArray[PosNum++];
if(NumArray[Range1] >= 1 && NumArray[Range1] <= 10){
Range1++;
}
if (NumArray[Range2] >= 11 && NumArray[Range2] <= 20){
Range2++;
}
if(NumArray[Range3] >= 21 && NumArray[Range3] <= 30){
Range3++;
}
if(NumArray[Range4] >= 31 && NumArray[Range4] <= 40){
Range4++;
}
if(NumArray[Range5] >= 41 && NumArray[Range5] <= 50){
Range5++;
}
if(NumArray[Range6] >= 51 && NumArray[Range6] <= 54){
Range6++;
}
}
Average = Total / 301;
System.out.println("The Average Is: " + Average);
System.out.println("1 Through 10 appear: " + Range1 + " Times");
System.out.println(" 11 Through 20 appear: " + Range2 + "Times");
System.out.println("21 Through 30 appear: " + Range3 + " Times");
System.out.println("31 Through 40 appear: " + Range4 + " Times");
System.out.println("41 Through 50 appear: " + Range5 + " Times");
System.out.println("51 Through 54 appear: " + Range6 + " Times");
}
}
这是我最新的尝试。
import java.util.*;
import java.io.*;
public class FixedInstream {
public static void main(String[] args) {
int [] MyArray = new int[301];
int PosNum = 0,
Average,
Total = 0,
R1 = 0,
R2 = 0,
R3 = 0,
R4 = 0,
R5 = 0,
R6 = 0,
R7 = 0;
try{
File Fout = new File("F:\\Text.txt");
PosNum = 0;
Scanner Fin = new Scanner(Fout);
while(Fin.hasNextInt()){
MyArray[PosNum] = Fin.nextInt();
Total += MyArray[PosNum];
}
Fin.close();
}catch(Exception E){
E.printStackTrace();
System.out.println("***ERROR ***** " + E.getMessage());
}
for(MyArray[PosNum] = 0; MyArray[PosNum] <= 300; ++MyArray[PosNum]){
if(MyArray[R1] >= 1 && MyArray[R1] <= 10){
R1++;
}
if (MyArray[R2] >= 11 && MyArray[R2] <= 20){
R2++;
}
if(MyArray[R3] >= 21 && MyArray[R3] <= 30){
R3++;
}
if(MyArray[R4] >= 31 && MyArray[R4] <= 40){
R4++;
}
if(MyArray[R5] >= 41 && MyArray[R5] <= 50){
R5++;
}
if(MyArray[R6] >= 51 && MyArray[R6] <= 54){
R6++;
}
}
Average = Total / 301;
System.out.println("The Average Is: " + Average);
System.out.println("1 Through 10 appear: " + R1 + " Times");
System.out.println("11 Through 20 appear: " + R2 + " Times");
System.out.println("21 Through 30 appear: " + R3 + " Times");
System.out.println("31 Through 40 appear: " + R4 + " Times");
System.out.println("41 Through 50 appear: " + R5 + " Times");
System.out.println("T51 Through 54 appear: " + R6 + " Times");
}
}
程序未正确输出下面的示例,我显然忘记在最初的评论中添加。
输出:
The Average Is: 27
The Numbers 1 Through 10 appear: 1 Times
The Numbers 11 Through 20 appear: 1 Times
The Numbers 21 Through 30 appear: 1 Times
The Numbers 31 Through 40 appear: 1 Times
The Numbers 41 Through 50 appear: 1 Times
The Numbers 51 Through 54 appear: 1 Times
文本文件 1 - 54 中有 300 个数字,这些数字重复的方式不同,但通常重复 10 - 50 次,正如您从上面的输出中看到的那样,它只显示一次,这让我相信我没有增加某些东西正确。
最佳答案
我认为这里有很多令人困惑的事情...下面是答案,但我建议您尝试自己调试此代码片段以了解问题所在。
您正在使用
301
迭代数组细胞。数组索引从0
开始至length-1
。当您使用numArray[++posNum]
进行迭代时当 posNum 初始化为 0 时,第一个索引将为++0 = 1
。所以你永远不会到达数组的第一个单元格。这导致out of range exception
因为最后一个索引将是301
但实际上最后一个索引是300
...只需阅读异常消息即可找到解决方案。我建议您看看
for loop
的工作原理了解如何使用递增索引逐单元浏览数组。然后你将像这样纠正你的 for 循环:for(int idx = 0; idx < NumArray.length; idx++)
您使用错误的变量访问当前单元格。这可以附加,但您会在 Debug模式下非常快地检测到它...
NumArray[idx]
而不是
NumArray[Range1]
这对于让你的程序正常工作并不是必需的,但是在 Java 中有一些约定(就像在每种语言中一样)和
variables
写在camelCase
.公共(public)类 InStream {
public static void main(String[] args) { int [] NumArray = new int[301]; int PosNum = 0, Total = 0, Average = 0, Range1 = 0, Range2 = 0, Range3 = 0, Range4 = 0, Range5 = 0, Range6 = 0; try{ File Fout = new File("F:\\Text.txt"); PosNum = 0; Scanner Fin = new Scanner(Fout); while(Fin.hasNextInt()){ NumArray[PosNum++] = Fin.nextInt(); } Fin.close(); }catch(Exception E){ E.printStackTrace(); System.out.println("***ERROR ***** " + E.getMessage()); } for(int idx = 0; idx < NumArray.length; idx++){ Total += NumArray[idx]; if(NumArray[idx] >= 1 && NumArray[idx] <= 10){ Range1++; } if (NumArray[idx] >= 11 && NumArray[idx] <= 20){ Range2++; } if(NumArray[idx] >= 21 && NumArray[idx] <= 30){ Range3++; } if(NumArray[idx] >= 31 && NumArray[idx] <= 40){ Range4++; } if(NumArray[idx] >= 41 && NumArray[idx] <= 50){ Range5++; } if(NumArray[idx] >= 51 && NumArray[idx] <= 54){ Range6++; } } Average = Total / 301; System.out.println("The Average Is: " + Average); System.out.println(" 1 Through 10 appear: " + Range1 + " Times"); System.out.println("11 Through 20 appear: " + Range2 + " Times"); System.out.println("21 Through 30 appear: " + Range3 + " Times"); System.out.println("31 Through 40 appear: " + Range4 + " Times"); System.out.println("41 Through 50 appear: " + Range5 + " Times"); System.out.println("51 Through 54 appear: " + Range6 + " Times"); }
}
不要犹豫,调试你的程序,这是学习语言和享受 Java 的最佳方式
关于将文本文件的内容放入数组然后显示和操作数组内容的 Java 问题,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33657890/