NumberFormatException異常而將字符串轉換爲int

Question

import java.util.*; 
class abc 
{ 
public static void main(String[] args) 
{ 
    int sum=0,n=4,k=10,i,j; 
    char ch[] = new char[10]; 
    char ans[] = new char[19]; 
    Scanner sc = new Scanner(System.in); 
    try1 abc=new try1(); 
String a,b,c,d; 
a= sc.nextLine(); 
b= sc.nextLine(); 
c= sc.nextLine(); 
d= sc.nextLine(); 
int num1 = (int) Long.parseLong(a,16); 
int num2 = (int) Long.parseLong(b,16); 
int num3 = (int) Long.parseLong(c,16); 
int num4 = (int) Long.parseLong(d,16); 

sum = num1 + num2 + num3 + num4; 
String sumstart = Integer.toHexString(sum); 
System.out.println(sumstart); 
String temp=sumstart; 
ch = temp.toCharArray(); 
for (i=1, j=0;i<5;i++,j++) 
{ 
    ans[j] = ch[i]; 
    } 
    ans[j]='\0'; 
    System.out.println(ans); 
    String anssing = Integer.toHexString(ch[0]); 
     int num5 = (int) Long.parseLong(anssing,16); 
     String ans1 = new String(ans); 

     System.out.println("ans 1 "+ans1); 

     int num6 =abc.convert(ans1); 
     System.out.println("num 6 vaala "+num6); 
     int ans2 = num5 + num6; 
     String hex4 = Integer.toHexString(ans2); 
     System.out.println("hex4 vaala "+hex4); 

} 
} 
class try1 
{ 
    int convert(String a) 
    { 

    int num5 = Integer.parseInt("3f18",16); 
    System.out.println(num5); 
    return num5; 
    } 
} 

考慮任何輸入，例如：

8fc6 
8fc6 
8fc6 
8fc6 

輸出：

Numberformatexception 

我已經存儲在字符數組的十六進制數，然後我將它轉換到字符串，現在我試圖將其轉換爲一個整數，但我收到NumberFormatException。任何人都可以幫我解決它嗎？

Answer 1

試試這個，

import java.util.*; 
class abc 
{ 
public static void main(String[] args) 
{ 
    int sum=0,n=4,k=10,i,j; 
    char ch[] = new char[10]; 
    char ans[] = new char[19]; 
    Scanner sc = new Scanner(System.in); 
    try1 abc=new try1(); 
String a,b,c,d; 
a= sc.nextLine(); 
b= sc.nextLine(); 
c= sc.nextLine(); 
d= sc.nextLine(); 
int num1 = Integer.parseInt(a,16); 
int num2 = Integer.parseInt(b,16); 
int num3 = Integer.parseInt(c,16); 
int num4 = Integer.parseInt(d,16); 

sum = num1 + num2 + num3 + num4; 
String sumstart = Integer.toHexString(sum); 
System.out.println(sumstart); 
String temp=sumstart; 
ch = temp.toCharArray(); 
for (i=1, j=0;i<5;i++,j++) 
{ 
    ans[j] = ch[i]; 
    } 
    ans[j]='\0'; 
    System.out.println(ans); 
    String anssing = Integer.toHexString(ch[0]); 
     int num5 = (int) Long.parseLong(anssing,16); 
     String ans1 = new String(ans); 

     System.out.println("ans 1 "+ans1); 

     int num6 =abc.convert(ans1); 
     System.out.println("num 6 vaala "+num6); 
     int ans2 = num5 + num6; 
     String hex4 = Integer.toHexString(ans2); 
     System.out.println("hex4 vaala "+hex4); 

} 
} 
class try1 
{ 
    int convert(String a) 
    { 

    int num5 = Integer.parseInt("3f18",16); 
    System.out.println(num5); 
    return num5; 
    } 
} 

解釋：我已經加入Integer.parseInt解析字符串輸入，以避免NumberFormatException的