Develop Reference

counting alphabetic characters between 'a' and 'z'

I came across a code which checks whether a character is between 'a' and 'z' case insensitive. However, I don't understand what the line after that is doing which is:
alphabets[c - 'a']++;
Could someone please explain this code to me?
alphabets = new int[26];
for (int i = 0; i < str.length(); i++)
{
char c = str.charAt(i);
if ('a' <= c && c <= 'z')
{
alphabets[c - 'a']++; // what does this do?
}
}

This code counts the number of times every lower-case letter appears in the strings. alphabets is an array where the first (i.e., index 0) index holds the number of as, the second the amount of bs, etc.
Subtracting a from the character will produce the relative index, and then ++ will increment the counter for that letter.

A char in Java is just a small integer, 16 bits wide. Generally speaking, the values it holds are the values that Unicode [aside: Java does not represent characters as "ASCII"] assigns to characters, but fundamentally, chars are just integers. Thus 'a' is the integer 0x0061, which can also be written as 97.
So, if you have value in the range 'a' to 'z', you have a value in the range 97 to 122. Subracting 'a' (subtracting 97) puts it in the range 0 to 25, which is suitable for indexing the 26-element array alphabets.

How do I get the numerical value/position of a character in the alphabet (1-26) in constant time (O(1)) without using any built in method or function?

How do I get the numerical value/position of a character in the alphabet (1-26) in constant time (O(1)) without using any built in method or function and without caring about the case of the character?

If your compiler supports binary literals you can use
int value = 0b00011111 & character;
If it does not, you can use 31 instead of 0b00011111 since they are equivalent.
int value = 31 & character;
or if you want to use hex
int value = 0x1F & character;
or in octal
int value = 037 & character;
You can use any way to represent the value 31.
This works because in ASCII, undercase values are prefixed with 011, and uppercase 010 and then the binary equivalent of 1-26.
By using the bitmask of 00011111 and the AND operand, we covert the 3 most significant bits to zeros. This leaves us with 00001 to 11010, 1 to 26.

Adding to the very good (self) answer of Charles Staal.
Assuming ascii encoding following will work. Updated from the kind comment of Yves Daoust
int Get1BasedIndex(char ch) {
return ( ch | ('a' ^ 'A') ) - 'a' + 1;
}
This will make the character uppercase and change the index.
However a more readable solution (O(1)) is:
int Get1BasedIndex(char ch) {
return ('a' <= ch && ch <= 'z') ? ch - 'a' + 1 : ch - 'A' + 1;
}
One more solution that is constant time but requires some extra memory is:
static int cha[256];
static void init() {
int code = -1;
fill_n (&cha[0], &cha[256], code);
code = 1;
for(char s = 'a', l = 'A'; s <= 'z'; ++s, ++l) {
cha[s] = cha[l] = code++;
}
}
int Get1BasedIndex(char ch) {
return cha[ch];
}

We can get their ASCII values and then subtract from the starting character ASCII(a - 97, A - 65)
char ch = 'a';
if(ch >=65 && ch <= 90)//if capital letter
System.out.println((int)ch - 65);
else if(ch >=97 && ch <= 122)//if small letters
System.out.println((int)ch - 97);

Strictly speaking it is not possible to do it portably in C/C++ because there is no guarantee on the ordering of the characters.
This said, with a contiguous sequence, Char - 'a' and Char - 'A' obviously give you the position of a lowercase or uppercase letter, and you could write
Ord= 'a' <= Char && Char <= 'z' ? Char - 'a' :
('A' <= Char && Char <= 'Z' ? Char - 'A' : -1);
If you want to favor efficiency over safety, exploit the binary representation of ASCII codes and use the branchless
#define ToUpper(Char) (Char | 0x20)
Ord= ToUpper(Char) - 'a';
(the output for non-letter character is considered unspecified).
Contrary to the specs, these snippets return the position in range [0, 25], more natural with zero-based indexing languages.

Converting characters to integers in Java

Can someone please explain to me what is going on here:
char c = '+';
int i = (int)c;
System.out.println("i: " + i + " ch: " + Character.getNumericValue(c));
This prints i: 43 ch:-1. Does that mean I have to rely on primitive conversions to convert char to int? So how can I convert a Character to Integer?
Edit: Yes I know Character.getNumericValue returns -1 if it is not a numeric value and that makes sense to me. The question is: why does doing primitive conversions return 43?
Edit2: 43 is the ASCII for +, but I would expect the cast to not succeed just like getNumericValue did not succeed. Otherwise that means there are two semantic equivalent ways to perform the same operation but with different results?

Character.getNumericValue(c)
The java.lang.Character.getNumericValue(char ch) returns the int value that the specified Unicode character represents. For example, the character '\u216C' (the roman numeral fifty) will return an int with a value of 50.
The letters A-Z in their uppercase ('\u0041' through '\u005A'), lowercase ('\u0061' through '\u007A'), and full width variant ('\uFF21' through '\uFF3A' and '\uFF41' through '\uFF5A') forms have numeric values from 10 through 35. This is independent of the Unicode specification, which does not assign numeric values to these char values.
This method returns the numeric value of the character, as a
nonnegative int value;
-2 if the character has a numeric value that is not a nonnegative integer;
-1 if the character has no numeric value.
And here is the link.

As the documentation clearly states, Character.getNumericValue() returns the character's value as a digit.
It returns -1 if the character is not a digit.
If you want to get the numeric Unicode code point of a boxed Character object, you'll need to unbox it first:
int value = (int)c.charValue();

Try any one of the below. These should work:
int a = Character.getNumericValue('3');
int a = Integer.parseInt(String.valueOf('3');

From the Javadoc for Character#getNumericValue:
If the character does not have a numeric value, then -1 is returned.
If the character has a numeric value that cannot be represented as a
nonnegative integer (for example, a fractional value), then -2 is
returned.
The character + does not have a numeric value, so you're getting -1.
Update:
The reason that primitive conversion is giving you 43 is that the the character '+' is encoded as the integer 43.

43 is the dec ascii number for the "+" symbol. That explains why you get a 43 back.
http://en.wikipedia.org/wiki/ASCII

public class IntergerParser {
public static void main(String[] args){
String number = "+123123";
System.out.println(parseInt(number));
}
private static int parseInt(String number){
char[] numChar = number.toCharArray();
int intValue = 0;
int decimal = 1;
for(int index = numChar.length ; index > 0 ; index --){
if(index == 1 ){
if(numChar[index - 1] == '-'){
return intValue * -1;
} else if(numChar[index - 1] == '+'){
return intValue;
}
}
intValue = intValue + (((int)numChar[index-1] - 48) * (decimal));
System.out.println((int)numChar[index-1] - 48+ " " + (decimal));
decimal = decimal * 10;
}
return intValue;
}

Why doesn't my compare work between char and int in Java?

char c = '0';
int i = 0;
System.out.println(c == i);
Why does this always returns false?

Although this question is very unclear, I am pretty sure the poster wants to know why this prints false:
char c = '0';
int i = 0;
System.out.println(c == i);
The answer is because every printable character is assigned a unique code number, and that's the value that a char has when treated as an int. The code number for the character 0 is decimal 48, and obviously 48 is not equal to 0.
Why aren't the character codes for the digits equal to the digits themselves? Mostly because the first few codes, especially 0, are too special to be used for such a mundane purpose.

The char c = '0' has the ascii code 48. This number is compared to s, not '0'. If you want to compare c with s you can either do:
if(c == s) // compare ascii code of c with s
This will be true if c = '0' and s = 48.
or
if(c == s + '0') // compare the digit represented by c
// with the digit represented by s
This will be true if c = '0' and s = 0.

The char and int value can not we directly compare we need to apply casting. So need to casting char to string and after string will pars into integer
char c='0';
int i=0;
Answer is like
String c = String.valueOf(c);
System.out.println(Integer.parseInt(c) == i)
It will return true;
Hope it will help you
Thanks

You're saying that s is an Integer and c (from what I see) is a Char.. so there you, that's the problem: Integer vs. Char comparation.

Conversion from ASCII values to Char

String source = "WEDGEZ"
char letter = source.charAt(i);
shift=5;
for (int i=0;i<source.length();i++){
if (source.charAt(i) >=65 && source.charAt(i) <=90 )
letterMix =(char)(('D' + (letter - 'D' + shift) % 26));
}
Ok what I'm trying to do is take the string WEDGEZ, and shift each letter by 5, so W becomes B and E becomes J, etc. However I feel like there is some inconsistency with the numbers I'm using.
For the if statement, I'm using ASCII values, and for the
letterMix= statement, I'm using the numbers from 1-26 (I think). Well actually, the question is about that too:
What does
(char)(('D' + (letter - 'D' + shift) % 26)); return anyway? It returns a char right, but converted from an int. I found that statement online somewhere I didn't compose it entirely myself so what exactly does that statement return.
The general problem with this code is that for W it returns '/' and for Z it returns _, which I'm guessing means it's using the ASCII values. I really dont know how to approach this.
Edit: New code
for (int i=0;i<source.length();i++)
{
char letter = source.charAt(i);
letterMix=source.charAt(i);
if (source.charAt(i) >=65 && source.charAt(i) <=90 ){
letterMix=(char)('A' + ( ( (letter - 'A') + input ) % 26));
}
}

Well I'm not sure if this homework, so i'll be stingy with the Code.
You're Writing a Caesar Cipher with a shift of 5.
To address your Z -> _ problem...I'm Assuming you want all the letters to be changed into encoded letters (and not weird Symbols). The problem is ASCII values of A-Z lie between 65 and 90.
When coding Z (for eg), you end up adding 5 to it, which gives u the value 95 (_).
What you need to do is Wrap around the available alphabets. First isolate, the relative position of the character in the alphabets (ie A = 0, B = 1 ...) You Need to subtract 65 (which is ASCII of A. Add your Shift and then apply modulus 26. This will cause your value to wrap around.
eg, it your encoding Z, (ASCII=90), so relative position is 25 (= 90 - 65).
now, 25 + 5 = 30, but you need the value to be within 26. so you take modulus 26
so 30 % 26 is 4 which is E.
So here it is
char letter = message(i);
int relativePosition = letter - 'A'; // 0-25
int encode = (relativePosition + shift) % 26
char encodedChar = encode + 'A' // convert it back to ASCII.
So in one line,
char encodedChar = 'A' + ( ( (letter - 'A') + shift ) % 26)
Note, This will work only for upper case, if your planning to use lower case, you'll need some extra processing.
You can use Character.isUpperCase() to check for upper case.

You can try this code for convert ASCII values to Char
class Ascii {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
char ch=sc.next().charAt(0);
if(ch==' ') {
int in=ch;
System.out.println(in);
}
}
}