测试时(使用以下代码)
char 'a'
是
10
,以及的ASCII值
char 'b'
是
11
。
然而
,连接时
字符“a”
和
字符'b'
,结果是
195
。
我在这里的逻辑和/或理解肯定有错误。。。我
做
理解字符不能连接为字符串,但是
195
然后代表?
这样的结果有什么用?
这是我的代码:
public class Concatenated
{
public static void main(String[] args)
{
char char1 = 'a';
char char2 = 'b';
String str1 = "abc";
String result = "";
int intResult = 0;
Concatenated obj = new Concatenated();
// calling methods here
intResult = obj.getASCII(char1);
System.out.println("The ASCII value of char \"" + char1 + "\" is: " + intResult + ".");
intResult = obj.getASCII(char2);
System.out.println("The ASCII value of char \"" + char2 + "\" is: " + intResult + ".");
result = obj.concatChars(char1, char2);
System.out.println(char1 + " + " + char2 + " = " + result + ".");
result = obj.concatCharString(char1, str1);
System.out.println("The char \"" + char1 + "\" + the String \"" + str1 + "\" = " + result + ".");
} // end of main method
public int getASCII(char testChar)
{
int ans = Character.getNumericValue(testChar);
return ans;
} // end of method getASCII
public String concatChars(char firstChar, char secondChar)
{
String ans = "";
ans += firstChar + secondChar; // "+=" is executed last
return ans; // returns ASCII value
} // end of method concatChars
public String concatCharString(char firstChar, String str)
{
String ans = "";
ans += firstChar + str;
return ans;
} // end of method concatCharString
} // end of class Concatenated
…打印到屏幕上的结果如下:
The ASCII value of char "a" is: 10.
The ASCII value of char "b" is: 11.
a + b = 195.
The char "a" + the String "abc" = aabc.
--编辑:--
正如@Marko Topolnik在下面指出的那样
getASCII
应更改为该值以返回正确的ASCII值(以及
不
UNICODE值):
public int getASCII(char testChar)
{
// int ans = Character.getNumericValue(testChar); ...returns a UNICODE value!
int ans = testChar;
return ans;
} // end of method getASCII
为了子孙后代,我没有改变上面的代码来反映这一点。