我最近发布了一个关于扫描仪未提供预期结果的问题，并了解到我的问题是我没有用.nextLine()刷新扫描仪。我对正在处理的程序感到困惑，因为我正在正确地刷新扫描仪，但是当我测试我的程序时，如果 - 当提示输入数字时 - 我输入一个字符串，我得到错误的输出。它重复两次相同的循环。
循环的顶部调用nextLine（），而处理无效输入的else块（例如我输入的字符串）也调用nextLine（）。但仍然不知怎的，我的输出结果不好
所以具体来说，这是一个错误输出的示例，用户输入为粗体，有问题的输出用斜体
输入左侧值： 2
输入运营商： -
输入右侧值： t
输入无效
输入运算符（+ - *或/：
无效的运营商
输入运算符（+ - *或/：
上面的四行会自动吐出到控制台。
以下是代码片段，其中包含错误代码的重要注释。我本来只发布了问题所在的while块，但由于while块是大部分程序而且整个程序只比该部分大一点，我认为发布所有内容会更好。
import java.util.Scanner;
import javax.swing.JOptionPane;
public class Calculator{
public static void main(String[] args){
double leftHandVal = 0.0;
//Output Title & Instructions
System.out.print("!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!\n");
System.out.print("!\t\t\t\t!\n");
System.out.print( "!\t INSTRUCTIONS\t\t!\n");
System.out.print("!\t\t\t\t!\n");
System.out.print("! INPUT\t\tOUTPUT\t\t!\n");
System.out.print("! *******\t *********\t\t!\n");
System.out.print("! c or C\t\tClear\t\t!\n");
System.out.print("! q or Q\t\tQuit\t\t!\n");
System.out.print("! +\t\tAddition\t\t!\n");
System.out.print("! -\t\tSubtraction\t!\n");
System.out.print("! *\t\tMultiplication\t!\n");
System.out.print("! /\t\tDivision\t\t!\n");
System.out.print("!\t\t\t\t!\n");
System.out.print("!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!\n\n");
while(true){
Scanner input = new Scanner(System.in);
char op = '\n';//(+, -, *, or /) will use in switch statement for their ascii decimal values
System.out.print("Enter the left-hand value: ");
//these blocks allow the code at the very bottom to not erroneously ask the user for extra input with hasNext() calls
if(input.hasNext("c") || input.hasNext("C")){//even though its unlikely for a user to clear so early...just in case
leftHandVal = 0.0;
}
else if(input.hasNext("q") || input.hasNext("Q")){//even though its unlikely for a user to quit so early...just in case
op = 113;//assign q for quit code
}
else if(input.hasNextDouble()){
leftHandVal = input.nextDouble();
/*
*
*BAD CODE INSIDE WHILE BELOW
*BAD CODE INSIDE WHILE BELOW
*BAD CODE INSIDE WHILE BELOW
*BAD CODE INSIDE WHILE BELOW
*
*/
while(true){
input.nextLine();
double rightHandVal = 0.0;
System.out.print("\nEnter operator (+ - * or / : ");
if(input.hasNext()){
op = input.next().charAt(0);
}
//if user wishes to cancel or quit on operator prompt, break out of inner while to access the clear and quit code
if(op == 99 || op == 67){
op = 99;
break;
}
else if(op == 113 || op == 81){
op = 113;
break;
}
else if((op != 43) && (op != 45) && (op != 42) && (op != 47)){//if invalid operator, restart inner while
System.out.print("Invalid Operator");
continue;
}
System.out.print("Enter the right-hand value: ");
if(input.hasNextDouble()){
rightHandVal = input.nextDouble();
switch(op){
case 43:
System.out.printf("%.3f + %.3f = %.3f", leftHandVal, rightHandVal, (leftHandVal + rightHandVal));
leftHandVal += rightHandVal;
break;
case 45:
System.out.printf("%.3f - %.3f = %.3f", leftHandVal, rightHandVal, (leftHandVal - rightHandVal));
leftHandVal -= rightHandVal;
break;
case 42:
System.out.printf("%.3f * %.3f = %.3f", leftHandVal, rightHandVal, (leftHandVal * rightHandVal));
leftHandVal *= rightHandVal;
break;
case 47:
System.out.printf("%.3f / %.3f = %.3f", leftHandVal, rightHandVal, (leftHandVal / rightHandVal));
leftHandVal /= rightHandVal;
break;
}
}
//if clear or quit requested from prompt for right-hand value, break to reach the clear and quit code
else if(input.hasNext("c") || input.hasNext("C")){
op = 99;
break;
}
else if(input.hasNext("q") || input.hasNext("Q")){
op = 113;
break;
}
else{
System.out.print("Invalid Input");
}
}
}
//if c || C reset op to null and restart outer while
if(op == 99 || op == 67){
op = '\n';
leftHandVal = 0.0;
continue;
}
//else if q || Q, prompt user with a popup to confirm.
if(op == 113 || op == 81){
int response = JOptionPane.showConfirmDialog(null, "QUIT CALCULATOR?", null, JOptionPane.YES_NO_OPTION);
if(response == 0){
System.exit(0);
}
continue;
}
}
}
}
答案 0 :(得分：1)
else if(input.hasNext("c") || input.hasNext("C"))
{
op = 99;
break;
}
else if(input.hasNext("q") || input.hasNext("Q")){
op = 113;
break;
}
else{
System.out.print("Invalid Input");
}
在此代码中，当您为右手值输入“t”时，您只需检查hasNext();，然后检查其他并输出无效输入。
但输入仍然具有值“t”，因此它会再次启动第二个而循环并且
System.out.print("\nEnter operator (+ - * or / : ");
if(input.hasNext()){
op = input.next().charAt(0);
}
检查已经有“t”的 input.hasNext（），所以取“t”并继续。
解决方法是在进入while循环之前刷新“t”。