将 String 数组转换为 int 值并将其存储在 int 数组中

Question

我有这部分代码：

int[] myIntArray = {0x2e80,0x008c,0x0993,0x09c5,0x058b,0x4c9c,0x0390,0x1e96,0x0989,0x0ac4,0x4cad,0x0d93,0x09c5,0x0a84,0x0591,0x04c5,0x058b,0x4c9c,0x0390,0x1ec5,0x0d87,0x0589,0x0591,0x0580,0x1fc4,0x4cb2,0x0591,0x048a,0x1991,0x4c84,0x4c8d,0x1988,0x0e89,0x09c5,0x0e90,0x18c5,0x1e80,0x0d96,0x038b,0x0d87,0x0080,0x4c86,0x038b,0x0a8c,0x0880,0x0286,0x09c5,0x058b,0x4c9c,0x0390,0x1ec5,0x0392,0x02c5,0x1c8a,0x1b80,0x1e96,0x4c9c,0x0390,0x4c86,0x0d8b,0x028a,0x18c5,0x0e80,0x4c96,0x1986,0x0f80,0x1f96,0x0a90,0x00c5,0x0397,0x4c8d,0x0d95,0x1c9c,0x42e5};
for (int i=0; i<=73; i++){
  String s1=Decrypt(k,myIntArray[i]);
  String s2= s1.substring(2,6);
  String s=convertHexToString(s2);
  System.out.print(s);
}

从数组中获取十六进制值并对其进行一些操作。而且它工作得很好，就像我想要的那样。

我想做同样的事情，但我想从文件中读取值并对其执行相同的操作，我试过了：

String token1 = "";
Scanner inFile1 = new Scanner(new    File("chipertext.txt")).useDelimiter(",\s*");
List<String> temps = new ArrayList<String>();
while (inFile1.hasNext()) {
  token1 = inFile1.next();
  temps.add(token1);
}
inFile1.close();
String[] tempsArray = temps.toArray(new String[74]);
int[] myIntArray = new int[tempsArray.length];

for (int i = 0; i < tempsArray.length; i++) {
  myIntArray[i] = Integer.parseInt(tempsArray[i]);
}
for (int i=0; i<=73; i++){
  String s1=Decrypt(k,myIntArray[i]);
  String s2= s1.substring(2,6);
  String s=convertHexToString(s2);
  System.out.print(s);
}

但是我得到这个错误：

Exception in thread "main" java.lang.NumberFormatException: For input string: "0x2e80  0x2e80
0x008c
0x0993
0x09c5
0x058b
0x4c9c
0x0390
0x1e96
0x0989
0x0ac4
0x4cad
0x0d93
0x09c5
0x0a84
0x0591
0x04c5
0x058b
0x4c9c
0x0390
0x1ec5
0x0d87
0x0589
0x0591
0x0580"

像这样存储在文件中的值：

0x2e80
0x008c
0x0993
0x09c5
0x058b
0x4c9c
0x0390
0x1e96
0x0989
0x0ac4
0x4cad
0x0d93
0x09c5
0x0a84
0x0591
0x04c5
0x058b
0x4c9c
0x0390
0x1ec5
0x0d87
0x0589
0x0591
0x0580

我认为这意味着该字符串不能存储为整数，对吗？那怎么办呢？以及它之前是如何存储在整数数组中的？！我不知道有人可以帮助我吗？

工作代码

String token1 = "";
Scanner inFile1 = new Scanner(new File("chipertext.txt"));
List<String> temps = new ArrayList<String>();
while (inFile1.hasNext()) {
token1 = inFile1.next();
temps.add(token1);
}
inFile1.close();
String[] tempsArray = temps.toArray(new String[73]);
int[] myIntArray = new int[tempsArray.length];
for (int i = 0; i < tempsArray.length; i++) {
myIntArray[i] = Integer.parseInt(tempsArray[i].substring(2), 16);
}
for (int i=0; i<=73; i++){
   String s1=Decrypt(k,myIntArray[i]);
   String s2= s1.substring(2,6);
   String s=convertHexToString(s2);
   System.out.print(s);
}

谢谢大家的帮助！！！

Answer 1

您发布的代码有两种可能性

你在文件中存储的十六进制代码不同

// 而不是
0x2e80
// 作为
2e80

// 并修改代码以指定基数
myIntArray[i] = Integer.parseInt(tempsArray[i], 16);
或将它们存储为当前

// 并修改代码去掉前缀指定基数
myIntArray[i] = Integer.parseInt(tempsArray[i].substring(2), 16);

查看两种解决方案的代码片段

对于案例 1。）

List<String> temps = new ArrayList<>();
temps.add("2e80");
String[] tempsArray = temps.toArray(new String[74]);
int parseInt = Integer.parseInt(tempsArray[0], 16);
System.out.println("parseInt = " + parseInt);

对于案例 2。）

List<String> temps = new ArrayList<>();
temps.add("0x2e80");
String[] tempsArray = temps.toArray(new String[74]);
int parseInt = Integer.parseInt(tempsArray[0].substring(2), 16);
System.out.println("parseInt = " + parseInt);

两个片段的输出是

parseInt = 11904

编辑

OP 代码的固定版本

// import java.nio.charset.Charset;
// import java.nio.file.Files;
// import java.nio.file.Paths;
// import java.util.List;

List<String> lines = Files.readAllLines(Paths.get("chipertext.txt"), 
        Charset.defaultCharset());
int[] myIntArray = new int[lines.size()];
for (int i = 0; i < myIntArray.length; i++) {
    myIntArray[i] = Integer.parseInt(lines.get(i).substring(2), 16);
}

Answer 2

如果您的所有号码格式相同，这将有效

String[] tempsArray = temps.toArray(new String[74]);
int[] myIntArray = new int[tempsArray.length];
for (int i = 0; i < tempsArray.length; i++) {
  myIntArray[i] = Integer.parseInt(tempsArray[i].substring(2), 16);
  System.out.println(myIntArray[i]);
}

如果它不起作用，则打印出进入循环的内容，看看您是否正确解析了文件。从您从解析的文件中获得的字符串数组中打印一项，并将其输入其中。

    String hex = "0x4c9c";
    int value = Integer.parseInt(hex.substring(2), 16);
    System.out.println(value);

打印出文件解析的输出，很可能它与您的预期不符。

List<String> temps = new ArrayList<String>();
while (inFile1.hasNext()) {
  token1 = inFile1.next();
  temps.add(token1);

  System.out.println(token1); //Check this output. Is it a hex string?
}
inFile1.close();

Answer 3

在这里查看 Tomasz Nurkiewicz 的回答：Parsing a Hexadecimal String to an Integer throws a NumberFormatException?

要恢复你需要使用

Integer.parseInt(tempsArray[i], 16)

设置为 16 的基数参数告诉 parseInt 解析 String 以获得十六进制值。在此之前，您需要删除每个“0x”

Answer 4

假设您正在存储 0x 以将其表示为十六进制数，您可以使用接受 radix 参数的 Integer.parseInt 方法。这里 radix 10 表示它的小数，radix 16 表示十六进制。并且您可以使用字符串的子字符串方法来删除 0x 噪音 .

所以解决方案如下所示

 System.out.println(Integer.parseInt("-FF", 16) ); // prints -255 

        String [] hexaDecimalNumbers = {"0x2e80","0x008c","0x0993","0x09c5","0x058b","0x4c9c","0x0390","0x1e96","0x0989","0x0ac4","0x4cad","0x0d93","0x09c5","0x0a84","0x0591","0x04c5","0x058b","0x4c9c","0x0390","0x1ec5","0x0d87","0x0589","0x0591","0x0580","0x1fc4","0x4cb2","0x0591","0x048a","0x1991","0x4c84","0x4c8d","0x1988","0x0e89","0x09c5","0x0e90","0x18c5","0x1e80","0x0d96","0x038b","0x0d87","0x0080","0x4c86","0x038b","0x0a8c","0x0880","0x0286","0x09c5","0x058b","0x4c9c","0x0390","0x1ec5","0x0392","0x02c5","0x1c8a","0x1b80","0x1e96","0x4c9c","0x0390","0x4c86","0x0d8b","0x028a","0x18c5","0x0e80","0x4c96","0x1986","0x0f80","0x1f96","0x0a90","0x00c5","0x0397","0x4c8d","0x0d95","0x1c9c","0x42e5"};
        for(String hexaDecimal: hexaDecimalNumbers) {
            System.out.print(Integer.parseInt(hexaDecimal.substring(2), 16) + " , ");
        }

当我将运行它打印输出如下

11904 , 140 , 2451 , 2501 , 1419 , 19612 , 912 , 7830 , 2441 , 2756 , 19629 , 3475 , 2501 , 2692 , 1425 , 1221 , 1419 , 19612 , 912 , 7877 , 3463 , 1417 , 1425 , 1408 , 8132 , 19634 , 1425 , 1162 , 6545 , 19588 , 19597 , 6536 , 3721 , 2501 , 3728 , 6341 , 7808 , 3478 , 907 , 3463 , 128 , 19590 , 907 , 2700 , 2176 , 646 , 2501 , 1419 , 19612 , 912 , 7877 , 914 , 709 , 7306 , 7040 , 7830 , 19612 , 912 , 19590 , 3467 , 650 , 6341 , 3712 , 19606 , 6534 , 3968 , 8086 , 2704 , 197 , 919 , 19597 , 3477 , 7324 , 17125 ,

Answer 5

而不是 Integer.parseInt(tempsArray[i]); 使用 Integer.parseInt(tempsArray[i].substring(2), 16));

将 String 数组转换为 int 值并将其存储在 int 数组中

Convert String array to int values and store it in int array

java

arrays

string

int

file