英文:
Convert a String Value to ASCII
问题
我想获取一个ASCII值（保存为字符串）并将其转换为字符以显示消息。我尝试了这个方法，但在声明int b时一直报告索引超出范围的错误。它还显示str和b没有值。
String value = "104 101 108 108 111";char[] ch = new char[value.length()];for (int i = 0; i < value.length(); i++) {ch[i] = value.charAt(i);}System.out.println(ch.length);String ans = "";int i = 0;while (i+2 < ch.length) {int b = ch[i] + ch[i+1] + ch[i+2];String str = new Character((char) b).toString();System.out.println(str);System.out.println(b);ans = ans + str;i = i + 3;}
英文:
I want to take the value of a ASCII value(Saved as a string) and convert it to the character to reveal a message. I tried this and it keeps throwing an index out of bound at the declaration of the int b.It also shows that str and b do not have a value
String value = "104 101 108 108 111";
char[] ch = new char[value.length()];for (int i = 0; i < value.length(); i++) {ch[i] = value.charAt(i);}System.out.println(ch.length);String ans = "";int i = 0;while (i+2 < ch.length) {int b= ch[i]+ch[i++]+ch[i+2];String str = new Character((char) b).toString();System.out.println(str);System.out.println(b);ans = ans+str;i=i+3;}
答案1
得分: 2
使用字符串分割函数
String value = "104 101 108 108 111";String[] arrOfStr = value.split(" ");String ans = "";for(String str : arrOfStr) {String str1 = Character.toString((char)Integer.parseInt(str));ans += str1;}System.out.println(ans); // 输出：hello
英文:
Using string split function
String value = "104 101 108 108 111";String[] arrOfStr = value.split(" ");String ans = "";for(String str : arrOfStr) {String str1 = Character.toString((char)Integer.parseInt(str));ans += str1;}System.out.println(ans); // output: hello
答案2
得分: 0
我们可以使用Java 8的Streams将“Imperative”代码切换为“Declarative”代码。
需要注意的关键点：
- 声明式风格更易读且更易编写。
- 字符串连接器比简单的字符串拼接更快。
- 无需编写迭代器。
import java.util.Arrays;public class Main {public static void main(String[] args) {String value = "104 101 108 108 111";Arrays.stream(value.split(" ")) // Starting a stream of String[].mapToInt(Integer::parseInt) // mapping String to int.mapToObj(Character::toChars) // finding ASCII char from int.forEach(System.out::print); // printing each character}}
如果您希望将结果存储起来然后打印出来，可以按照以下方式进行操作。
import java.util.Arrays;import java.util.stream.Collectors;public class Main {public static void main(String[] args) {String value = "104 101 108 108 111";String result = Arrays.stream(value.split(" ")) // Starting a stream of String[].mapToInt(Integer::parseInt) // mapping String to int.mapToObj(Character::toChars) // finding ASCII char from int.map(String::new) // convert char to String.collect(Collectors.joining()); // combining individual result using String JoinerSystem.out.println(result);}}
英文:
We can switch the Imperative code to Declarative code using Java 8 Streams.
Key points to observe:
- Declarative style is more readable and easy to write.
- String Joiner is faster than simple String Concatenation.
- No need to write an iterator.
import java.util.Arrays;public class Main {public static void main(String[] args) {String value = "104 101 108 108 111";Arrays.stream(value.split(" ")) // Starting a stream of String[].mapToInt(Integer::parseInt) // mapping String to int.mapToObj(Character::toChars) // finding ASCII char from int.forEach(System.out::print); // printing each character}}
If you wish to store the result and then print it, this is how is done.
import java.util.Arrays;import java.util.stream.Collectors;public class Main {public static void main(String[] args) {String value = "104 101 108 108 111";String result = Arrays.stream(value.split(" ")) // Starting a stream of String[].mapToInt(Integer::parseInt) // mapping String to int.mapToObj(Character::toChars) // finding ASCII char from int.map(String::new) // convert char to String.collect(Collectors.joining()); // combining individual result using String JoinerSystem.out.println(result);}}
答案3
得分: -1
// 代码注释：// 有一个内置方法可以获得字符串的字符数组 `char[]`，因此下面这两个代码块是相同的：```lang-java// 来自问题的代码char[] ch = new char[value.length()];for (int i = 0; i < value.length(); i++) {ch[i] = value.charAt(i);}// 使用内置方法char[] ch = value.toCharArray();
在循环时递增一个值时最好使用 for 循环。下面两种方式的循环行为相同，但 for 循环将循环逻辑放在一起：
// 来自问题的代码int i = 0;while (i+2 < ch.length) {// 这里有一些代码i=i+3;}// 使用 for 循环for (int i = 0; i + 2 < ch.length; i=i+3) {// 这里有一些代码}
下面这行代码是完全错误的：
int b= ch[i]+ch[i++]+ch[i+2];
-
i++递增了i的值，但在表达式中使用的是递增前的值，这意味着如果在此行之前i = 0，结果与以下代码相同：int b = ch[0] + ch[0] + ch[2];i = i + 1;
您需要用
i + 1替换i++，并意识到它们是不同的。 -
由于您不再在该语句中将
i的值递增 1，因此循环必须从i=i+3更改为i = i + 4，以正确跳过输入中的空格。 -
ch[i]的值是一个char值，通过使用+运算符将其扩展为int值。char的int值是 Unicode 代码点值，对于您的文本，它也是字符的 ASCII 码。这意味着如果
i = 0，在修复问题＃1后，表达式将计算为：int b = ch[0] + ch[1] + ch[2];int b = `1` + `0` + `4`;int b = 49 + 48 + 52;int b = 149;
这与在问题中运行代码的输出相匹配，其中第二个打印的数字是
149（在修复问题＃1后）。
您真正想要的是获取子字符串 "104"，将其转换为数字，然后将该 ASCII 码值强制转换为 char，就像这样：
String numberStr = value.substring(i, i + 3); // 例如："104"int number = Integer.parseInt(numberStr); // 例如：104String str = String.valueOf((char) number); // 例如："h"
有了这个，您就不再需要 char[]，因此最终的代码将是：
String value = "104 101 108 108 111";String ans = "";for (int i = 0; i + 2 < value.length(); i += 4) {String numberStr = value.substring(i, i + 3);int number = Integer.parseInt(numberStr);String str = String.valueOf((char) number);ans = ans + str;}System.out.println(ans);
输出
hello
英文:
Comments to code:
There is a built-in method for for getting a char[] with the characters of a string, so the following two blocks of code are the same:
// Code from questionchar[] ch = new char[value.length()];for (int i = 0; i < value.length(); i++) {ch[i] = value.charAt(i);}// Using built-in methodchar[] ch = value.toCharArray();
It is better to use a for loop when increment a value while looping. The following two ways of writing the loop behave the same, but the for loop keeps the loop logic together:
// Code from questionint i = 0;while (i+2 < ch.length) {// some code herei=i+3;}// Using for loopfor (int i = 0; i + 2 < ch.length; i=i+3) {// some code here}
The following line of code is entirely wrong:
int b= ch[i]+ch[i++]+ch[i+2];
-
i++increments the value isi, but it is the value before the increment that is used in the expression, which means that ifi = 0before the line, the result is the same as this code:int b = ch[0] + ch[0] + ch[2];i = i + 1;
You need to replace
i++withi + 1, and realize that those are not the same. -
Since you no longer increment the value of
iby 1 in that statement, the loop much be changed fromi=i+3toi = i + 4, to correctly skip the spaces in the input. -
The value of
ch[i]is acharvalue, which is widened to anintvalue by the use of the+operator. Theintvalue of acharis the Unicode Code Point value, which for your text is also the same as the ASCII code for the character.This means that if
i = 0, the expression would (after fixing issue #1) evaluate as:int b = ch[0] + ch[1] + ch[2];int b = `1` + `0` + `4`;int b = 49 + 48 + 52;int b = 149;
That matches the output from running the code is in question, where the second printed number is
149(after fixing issue #1).
What you really wanted was to get the substring "104" and convert that to a number, then cast that ASCII code value to a char, like this:
String numberStr = value.substring(i, i + 3); // E.g. "104"int number = Integer.parseInt(numberStr); // E.g. 104String str = String.valueOf((char) number); // E.g. "h"
With that, you no longer need the char[], so the final code would be:
String value = "104 101 108 108 111";String ans = "";for (int i = 0; i + 2 < value.length(); i += 4) {String numberStr = value.substring(i, i + 3);int number = Integer.parseInt(numberStr);String str = String.valueOf((char) number);ans = ans + str;}System.out.println(ans);
Output
hello
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