I have this:
char c = "\ud804\udef4".charAt(0);
char d = "\ud804\udef4".charAt(1);
How will I print c and d as hex Strings ?
I want d804 for c and def4 for d.
It doesn't matter whether the char is a surrogate pair or not. If you have a char, you can convert it to a hex string by Integer.toHexString(), since chars can be implicitly converted to int.
System.out.println(Integer.toHexString(c));
System.out.println(Integer.toHexString(d));
Related
String message = "a";
char message1 = (char) message;
System.out.println(message1);
Gives me an output error,
This should be converted with ease because the string is one character "a"
I know I can do it explicitly sorry, why the two are incompatible to cast if they are storing the same (only one character)?
As you've seen, no, you cannot cast a single character String to a char. But you could extract it explicitly:
String message = "a";
char message1 = message.charAt(0);
No you cannot do that. You can cast a char to Character because the Character object type is the "boxed" version of the char base type.
Character charObject = (Character) 'c';
char charBase = (char) charObject;
actually, because of auto-boxing and auto-unboxing, you don't need the explicit cast:
Character charObject = 'c';
char charBase = charObject;
However, a String is an object type much like any other object type. That means you cannot cast it to char, you need to use the charAt(int index) method to retrieve characters from it.
Beware though that you may want to use codePointAt(int index) instead, since Unicode code points may well extend out of the 65536 code points that can be stored in the 16 bits that a char represents. So please make sure that no characters defined in the "supplementary planes" are present in your string when using charAt(int index).
As in Java any type can be converted to String, it is possibly to directly append characters to a string though, so "strin" + 'g' works fine. This is also because the + operator for String is syntactic sugar in Java (i.e. other objects cannot use + as operator, you would have to use a method such as append()). Do remember that it returns a new string rather than expanding the original "strin" string. Java strings are immutable after all.
You cannot cast a String to a char. Below is a snippet to always pick the first character from the String,
char c = message.charAt(0);
In case you want to convert the String to a character array, then it can be done as,
String g = "test";
char[] c_arr = g.toCharArray(); // returns a length 4 char array ['t','e','s','t']
A String with one char is more akin to a char[1]. Regardless, retrieve the character directly:
String ex = /* your string */;
if (!ex.isEmpty()) {
char first = ex.charAt(0);
}
I want to know how to divide a String and seperate it into two variables, one as char and other as integer
Example:
If "C 365" Is my string then char variable will be 'C' and Integer variable will be 365
The char can be extracted with charAt. For the int, just use substring to take the string from the third character (the first is the char and the second is a space), and then parse it:
String str = "C 365";
char ch = str.charAt(0);
int i = Integer.parseInt(str.substring(2));
Is there a method of java characters that will return the next character in lexicographical order?
char x = 'a'.next();
'a'.next() would return b. 'b'.next() would return c. etc.
char x = 'a';
System.out.println(++x);
Output: b
I hope this is enough as a hint.
Basically, just increase the ASCII code of char variable to get the next char.
And yes, ASCII code is basically just an integer. (Try doing System.out.println((int) 'a'); and you will get output as 97, which is the ASCII value for char a). Similarly, every char has some ASCII value.
I looked for this about an hour now, but couldn't get any advise specific to my problem. What I'd like to do is take a string of 0's and 1's and manipulate a char that it fits the given String pattern. For example:
char c = 'b'
String s = "00000000 01100001";
Now I'd like to manipulate the bits in c, so that they match the bit pattern specified in s. As result c would be printed as 'a' (if I'm not completely wrong about it). Any help appreciated!
You can do
char a = (char) Integer.parseInt("0000000001100001", 2);
To do the conversion from binary string to Integer, use parseInt with the 2nd argument as 2.
int temp = Integer.parseInt("01100001", 2);
You can modify with binary operators (&,|,^), but if what you really want is to just assign a variable, you can do it with casts.
char c = 'c';
System.out.println((char)(c&temp));
System.out.println((char)temp);
How about:
String s = "00000000 01100001";
String[] w = s.split(" ");
char c = (char)(Integer.parseInt(w[0], 2) * 256 + Integer.parseInt(w[1], 2));
This allows for the leading zeroes of each byte to be omitted. If you know they're there, you can just replace the space out of the string and use a single parseInt() call:
char c = (char)Integer.parseInt(s.replace(" ", ""), 2);
So I just started reading "Java In A Nutshell", and on Chapter One it states that:
"To include a character literal in a Java program, simply place it between single quotes"
i.e.
char c = 'A';
What exactly does this do^? I thought char only took in values 0 - 65,535. I don't understand how you can assign 'A' to it?
You can also assign 'B' to an int?
int a = 'B'
The output for 'a' is 66. Where/why would you use the above^ operation?
I apologise if this is a stupid question.
My whole life has been a lie.
char is actually an integer type. It stores the 16-bit Unicode integer value of the character in question.
You can look at something like http://asciitable.com to see the different values for different characters.
In Java char literals represent UTF-16 (character encoding schema) code units. What you got from UTF-16 is mapping between integer values (and the way they are saved in memory) with corresponding character (graphical representation of unit code).
You can enclose characters in single quotes - this way you don't need to remember UTF-16 values for characters you use. You can still get the integer value from character type and put if for example in int type (but generally not in short, they both use 16 bits but short values are from -32768 to 32767 and char values are from 0 to 65535 or so).
If you look at an ASCII chart, the character "A" has a value of 41 hex or 65 decimal. Using the ' character to bracket a single character makes it a character literal. Using the double-quote (") would make it a String literal.
Assigning char someChar = 'A'; is exactly the same as saying char someChar = 65;.
As to why, consider if you simply want to see if a String contains a decimal number (and you don't have a convenient function to do this). You could use something like:
bool isDecimal = true;
for (int i = 0; i < decString.length(); i++) {
char theChar = decString.charAt(i);
if (theChar < '0' || theChar > '9') {
isDecimal = false;
break;
}
}