Converting and storing integers to ASCII Characters [duplicate] - java

How do you concatenate characters in java? Concatenating strings would only require a + between the strings, but concatenating chars using + will change the value of the char into ascii and hence giving a numerical output. I want to do System.out.println(char1+char2+char3... and create a String word like this.
I could do
System.out.print(char1);
System.out.print(char2);
System.out.print(char3);
But, this will only get me the characters in 1 line. I need it as a string. Any help would be appreciated.
Thanks

Do you want to make a string out of them?
String s = new StringBuilder().append(char1).append(char2).append(char3).toString();
Note that
String b = "b";
String s = "a" + b + "c";
Actually compiles to
String s = new StringBuilder("a").append(b).append("c").toString();
Edit: as litb pointed out, you can also do this:
"" + char1 + char2 + char3;
That compiles to the following:
new StringBuilder().append("").append(c).append(c1).append(c2).toString();
Edit (2): Corrected string append comparison since, as cletus points out, a series of strings is handled by the compiler.
The purpose of the above is to illustrate what the compiler does, not to tell you what you should do.

I wasn't going to answer this question but there are two answers here (that are getting voted up!) that are just plain wrong. Consider these expressions:
String a = "a" + "b" + "c";
String b = System.getProperty("blah") + "b";
The first is evaluated at compile-time. The second is evaluated at run-time.
So never replace constant concatenations (of any type) with StringBuilder, StringBuffer or the like. Only use those where variables are invovled and generally only when you're appending a lot of operands or you're appending in a loop.
If the characters are constant, this is fine:
String s = "" + 'a' + 'b' + 'c';
If however they aren't, consider this:
String concat(char... chars) {
if (chars.length == 0) {
return "";
}
StringBuilder s = new StringBuilder(chars.length);
for (char c : chars) {
s.append(c);
}
return s.toString();
}
as an appropriate solution.
However some might be tempted to optimise:
String s = "Name: '" + name + "'"; // String name;
into this:
String s = new StringBuilder().append("Name: ").append(name).append("'").toString();
While this is well-intentioned, the bottom line is DON'T.
Why? As another answer correctly pointed out: the compiler does this for you. So in doing it yourself, you're not allowing the compiler to optimise the code or not depending if its a good idea, the code is harder to read and its unnecessarily complicated.
For low-level optimisation the compiler is better at optimising code than you are.
Let the compiler do its job. In this case the worst case scenario is that the compiler implicitly changes your code to exactly what you wrote. Concatenating 2-3 Strings might be more efficient than constructing a StringBuilder so it might be better to leave it as is. The compiler knows whats best in this regard.

If you have a bunch of chars and want to concat them into a string, why not do
System.out.println("" + char1 + char2 + char3);
?

You can use the String constructor.
System.out.println(new String(new char[]{a,b,c}));

You need to tell the compiler you want to do String concatenation by starting the sequence with a string, even an empty one. Like so:
System.out.println("" + char1 + char2 + char3...);

System.out.println(char1+""+char2+char3)
or
String s = char1+""+char2+char3;

You need a String object of some description to hold your array of concatenated chars, since the char type will hold only a single character. e.g.,
StringBuilder sb = new StringBuilder('a').append('b').append('c');
System.out.println(sb.toString);

public class initials {
public static void main (String [] args) {
char initialA = 'M';
char initialB = 'P';
char initialC = 'T';
System.out.println("" + initialA + initialB + initialC );
}
}

I don't really consider myself a Java programmer, but just thought I'd add it here "for completeness"; using the (C-inspired) String.format static method:
String s = String.format("%s%s", 'a', 'b'); // s is "ab"

this is very simple approach to concatenate or append the character
StringBuilder desc = new StringBuilder();
String Description="this is my land";
desc=desc.append(Description.charAt(i));

simple example to selecting character from string and appending to string variable
private static String findaccountnum(String holdername, String mobile) {
char n1=holdername.charAt(0);
char n2=holdername.charAt(1);
char n3=holdername.charAt(2);
char n4=mobile.charAt(0);
char n5=mobile.charAt(1);
char n6=mobile.charAt(2);
String number=new StringBuilder().append(n1).append(n2).append(n3).append(n4).append(n5).append(n6).toString();
return number;
}

System.out.print(a + "" + b + "" + c);

Related

How to - Loop through a String and identify a specific character and add to the string

I'm currently trying to loop through a String and identity a specific character within that string then add a specific character following on from the originally identified character.
For example using the string: aaaabbbcbbcbb
And the character I want to identify being: c
So every time a c is detected a following c will be added to the string and the loop will continue.
Thus aaaabbbcbbcbb will become aaaabbbccbbccbb.
I've been trying to make use of indexOf(),substring and charAt() but I'm currently either overriding other characters with a c or only detecting one c.
I know you've asked for a loop, but won't something as simple as a replace suffice?
String inputString = "aaaabbbcbbcbb";
String charToDouble = "c";
String result = inputString.replace(charToDouble, charToDouble+charToDouble);
// or `charToDouble+charToDouble` could be `charToDouble.repeat(2)` in JDK 11+
Try it online.
If you insist on using a loop however:
String inputString = "aaaabbbcbbcbb";
char charToDouble = 'c';
String result = "";
for(char c : inputString.toCharArray()){
result += c;
if(c == charToDouble){
result += c;
}
}
Try it online.
Iterate over all the characters. Add each one to a StringBuilder. If it matches the character you're looking for then add it again.
final String test = "aaaabbbcbbcbb";
final char searchChar = 'c';
final StringBuilder builder = new StringBuilder();
for (final char c : test.toCharArray())
{
builder.append(c);
if (c == searchChar)
{
builder.append(c);
}
}
System.out.println(builder.toString());
Output
aaaabbbccbbccbb
You probably are trying to modify a String in java. Strings in Java are immutable and cannot be changed like one might do in c++.
You can use StringBuilder to insert characters. eg:
StringBuilder builder = new StringBuilder("acb");
builder.insert(1, 'c');
The previous answer suggesting String.replace is the best solution, but if you need to do it some other way (e.g. for an exercise), then here's a 'modern' solution:
public static void main(String[] args) {
final String inputString = "aaaabbbcbbcbb";
final int charToDouble = 'c'; // A Unicode codepoint
final String result = inputString.codePoints()
.flatMap(c -> c == charToDouble ? IntStream.of(c, c) : IntStream.of(c))
.collect(StringBuilder::new, StringBuilder::appendCodePoint, StringBuilder::append)
.toString();
assert result.equals("aaaabbbccbbccbb");
}
This looks at each character in turn (in an IntStream). It doubles the character if it matches the target. It then accumulates each character in a StringBuilder.
A micro-optimization can be made to pre-allocate the StringBuilder's capacity. We know the maximum possible size of the new string is double the old string, so StringBuilder::new can be replaced by () -> new StringBuilder(inputString.length()*2). However, I'm not sure if it's worth the sacrifice in readability.

Java StringBuilder character appending returns unwanted number

I am very sorry if this is a basic question which has been answered before (I tried looking but I did not find anything)
I am trying to write the following Java method:
String winningCard(String trick, char trump) {
StringBuilder sb = new StringBuilder();
char suit;
char rank;
for(int i = 1; i < trick.length(); i+=2) {
if(trick.charAt(i) == trump) {
suit = trick.charAt(i);
rank = trick.charAt(i-1);
sb.append(rank + suit); //issue here, returns a weird number
break;
}
}
String result = sb.toString();
return result;
}
When called with these arguments "8s7hQd", 'h' for example, it is supposed to return "7h".
If I change the StringBuilder to only append either the suit or the rank, it does it just fine, but if I put it the way it is above it returns "159" which I believe has something to do with the unicode encoding.
I'd very much appreciate if a kind sould could tell me what I am missing.
Thanks in advance
suit and rank are basically numbers. The + is adding these numbers and appending it.
If you place a "" between, the chars will be appended as you intend, because it forces the compiler to use the + with a String.
sb.append(rank + "" + suit);
append(rank).append(suit);
Should do the trick
+ is a tricksy thing, because it means different things in different contexts.
If at least one of the operands is a String, it acts as the string concatenation operator.
If both of the operands are numbers, or convertible to numbers via unboxing, then it acts as the numeric addition operator.
You are giving it two chars: these are numbers, so numeric addition occurs.
Before adding the two chars, they are widened to int; the result is an int too. And it is this int that you are appending to the string builder, hence the "unwanted" number.
So, either avoid using the addition operator at all (best):
sb.append(rank).append(suit);
Or make sure you are using the string concatenation operator:
sb.append("" + rank + suit);
// Left-associative, so evaluated as
// ("" + rank) + suit
sb.append(String.valueOf(rank) + suit);
// Etc.
But actually, you don't need to do either: just append the substring:
sb.append(trick, i-1, i+1);
This extracts a portion of the trick string, as trick.substring(i-1, i+1) would, but does it without creating a new string.
And you don't need a loop
You can say directly that those chars should be interpreted as String by
sb.append(String.valueOf(rank) + String.valueOf(suit))

Using only 1 System.out.print() instead of 3. More details below [duplicate]

I am trying to concatenate strings in Java. Why isn't this working?
public class StackOverflowTest {
public static void main(String args[]) {
int theNumber = 42;
System.out.println("Your number is " . theNumber . "!");
}
}
You can concatenate Strings using the + operator:
System.out.println("Your number is " + theNumber + "!");
theNumber is implicitly converted to the String "42".
The concatenation operator in java is +, not .
Read this (including all subsections) before you start. Try to stop thinking the php way ;)
To broaden your view on using strings in Java - the + operator for strings is actually transformed (by the compiler) into something similar to:
new StringBuilder().append("firstString").append("secondString").toString()
There are two basic answers to this question:
[simple] Use the + operator (string concatenation). "your number is" + theNumber + "!" (as noted elsewhere)
[less simple]: Use StringBuilder (or StringBuffer).
StringBuilder value;
value.append("your number is");
value.append(theNumber);
value.append("!");
value.toString();
I recommend against stacking operations like this:
new StringBuilder().append("I").append("like to write").append("confusing code");
Edit: starting in java 5 the string concatenation operator is translated into StringBuilder calls by the compiler. Because of this, both methods above are equal.
Note: Spaceisavaluablecommodity,asthissentancedemonstrates.
Caveat: Example 1 below generates multiple StringBuilder instances and is less efficient than example 2 below
Example 1
String Blam = one + two;
Blam += three + four;
Blam += five + six;
Example 2
String Blam = one + two + three + four + five + six;
Out of the box you have 3 ways to inject the value of a variable into a String as you try to achieve:
1. The simplest way
You can simply use the operator + between a String and any object or primitive type, it will automatically concatenate the String and
In case of an object, the value of String.valueOf(obj) corresponding to the String "null" if obj is null otherwise the value of obj.toString().
In case of a primitive type, the equivalent of String.valueOf(<primitive-type>).
Example with a non null object:
Integer theNumber = 42;
System.out.println("Your number is " + theNumber + "!");
Output:
Your number is 42!
Example with a null object:
Integer theNumber = null;
System.out.println("Your number is " + theNumber + "!");
Output:
Your number is null!
Example with a primitive type:
int theNumber = 42;
System.out.println("Your number is " + theNumber + "!");
Output:
Your number is 42!
2. The explicit way and potentially the most efficient one
You can use StringBuilder (or StringBuffer the thread-safe outdated counterpart) to build your String using the append methods.
Example:
int theNumber = 42;
StringBuilder buffer = new StringBuilder()
.append("Your number is ").append(theNumber).append('!');
System.out.println(buffer.toString()); // or simply System.out.println(buffer)
Output:
Your number is 42!
Behind the scene, this is actually how recent java compilers convert all the String concatenations done with the operator +, the only difference with the previous way is that you have the full control.
Indeed, the compilers will use the default constructor so the default capacity (16) as they have no idea what would be the final length of the String to build, which means that if the final length is greater than 16, the capacity will be necessarily extended which has price in term of performances.
So if you know in advance that the size of your final String will be greater than 16, it will be much more efficient to use this approach to provide a better initial capacity. For instance, in our example we create a String whose length is greater than 16, so for better performances it should be rewritten as next:
Example optimized :
int theNumber = 42;
StringBuilder buffer = new StringBuilder(18)
.append("Your number is ").append(theNumber).append('!');
System.out.println(buffer)
Output:
Your number is 42!
3. The most readable way
You can use the methods String.format(locale, format, args) or String.format(format, args) that both rely on a Formatter to build your String. This allows you to specify the format of your final String by using place holders that will be replaced by the value of the arguments.
Example:
int theNumber = 42;
System.out.println(String.format("Your number is %d!", theNumber));
// Or if we need to print only we can use printf
System.out.printf("Your number is still %d with printf!%n", theNumber);
Output:
Your number is 42!
Your number is still 42 with printf!
The most interesting aspect with this approach is the fact that we have a clear idea of what will be the final String because it is much more easy to read so it is much more easy to maintain.
The java 8 way:
StringJoiner sj1 = new StringJoiner(", ");
String joined = sj1.add("one").add("two").toString();
// one, two
System.out.println(joined);
StringJoiner sj2 = new StringJoiner(", ","{", "}");
String joined2 = sj2.add("Jake").add("John").add("Carl").toString();
// {Jake, John, Carl}
System.out.println(joined2);
You must be a PHP programmer.
Use a + sign.
System.out.println("Your number is " + theNumber + "!");
"+" instead of "."
Use + for string concatenation.
"Your number is " + theNumber + "!"
This should work
public class StackOverflowTest
{
public static void main(String args[])
{
int theNumber = 42;
System.out.println("Your number is " + theNumber + "!");
}
}
For exact concatenation operation of two string please use:
file_names = file_names.concat(file_names1);
In your case use + instead of .
For better performance use str1.concat(str2) where str1 and str2 are string variables.
String.join( delimiter , stringA , stringB , … )
As of Java 8 and later, we can use String.join.
Caveat: You must pass all String or CharSequence objects. So your int variable 42 does not work directly. One alternative is using an object rather than primitive, and then calling toString.
Integer theNumber = 42;
String output =
String // `String` class in Java 8 and later gained the new `join` method.
.join( // Static method on the `String` class.
"" , // Delimiter.
"Your number is " , theNumber.toString() , "!" ) ; // A series of `String` or `CharSequence` objects that you want to join.
) // Returns a `String` object of all the objects joined together separated by the delimiter.
;
Dump to console.
System.out.println( output ) ;
See this code run live at IdeOne.com.
In java concatenate symbol is "+".
If you are trying to concatenate two or three strings while using jdbc then use this:
String u = t1.getString();
String v = t2.getString();
String w = t3.getString();
String X = u + "" + v + "" + w;
st.setString(1, X);
Here "" is used for space only.
In Java, the concatenation symbol is "+", not ".".
"+" not "."
But be careful with String concatenation. Here's a link introducing some thoughts from IBM DeveloperWorks.
You can concatenate Strings using the + operator:
String a="hello ";
String b="world.";
System.out.println(a+b);
Output:
hello world.
That's it
So from the able answer's you might have got the answer for why your snippet is not working. Now I'll add my suggestions on how to do it effectively. This article is a good place where the author speaks about different way to concatenate the string and also given the time comparison results between various results.
Different ways by which Strings could be concatenated in Java
By using + operator (20 + "")
By using concat method in String class
Using StringBuffer
By using StringBuilder
Method 1:
This is a non-recommended way of doing. Why? When you use it with integers and characters you should be explicitly very conscious of transforming the integer to toString() before appending the string or else it would treat the characters to ASCI int's and would perform addition on the top.
String temp = "" + 200 + 'B';
//This is translated internally into,
new StringBuilder().append( "" ).append( 200 ).append('B').toString();
Method 2:
This is the inner concat method's implementation
public String concat(String str) {
int olen = str.length();
if (olen == 0) {
return this;
}
if (coder() == str.coder()) {
byte[] val = this.value;
byte[] oval = str.value;
int len = val.length + oval.length;
byte[] buf = Arrays.copyOf(val, len);
System.arraycopy(oval, 0, buf, val.length, oval.length);
return new String(buf, coder);
}
int len = length();
byte[] buf = StringUTF16.newBytesFor(len + olen);
getBytes(buf, 0, UTF16);
str.getBytes(buf, len, UTF16);
return new String(buf, UTF16);
}
This creates a new buffer each time and copies the old content to the newly allocated buffer. So, this is would be too slow when you do it on more Strings.
Method 3:
This is thread safe and comparatively fast compared to (1) and (2). This uses StringBuilder internally and when it allocates new memory for the buffer (say it's current size is 10) it would increment it's 2*size + 2 (which is 22). So when the array becomes bigger and bigger this would really perform better as it need not allocate buffer size each and every time for every append call.
private int newCapacity(int minCapacity) {
// overflow-conscious code
int oldCapacity = value.length >> coder;
int newCapacity = (oldCapacity << 1) + 2;
if (newCapacity - minCapacity < 0) {
newCapacity = minCapacity;
}
int SAFE_BOUND = MAX_ARRAY_SIZE >> coder;
return (newCapacity <= 0 || SAFE_BOUND - newCapacity < 0)
? hugeCapacity(minCapacity)
: newCapacity;
}
private int hugeCapacity(int minCapacity) {
int SAFE_BOUND = MAX_ARRAY_SIZE >> coder;
int UNSAFE_BOUND = Integer.MAX_VALUE >> coder;
if (UNSAFE_BOUND - minCapacity < 0) { // overflow
throw new OutOfMemoryError();
}
return (minCapacity > SAFE_BOUND)
? minCapacity : SAFE_BOUND;
}
Method 4
StringBuilder would be the fastest one for String concatenation since it's not thread safe. Unless you are very sure that your class which uses this is single ton I would highly recommend not to use this one.
In short, use StringBuffer until you are not sure that your code could be used by multiple threads. If you are damn sure, that your class is singleton then go ahead with StringBuilder for concatenation.
First method: You could use "+" sign for concatenating strings, but this always happens in print.
Another way: The String class includes a method for concatenating two strings: string1.concat(string2);
import com.google.common.base.Joiner;
String delimiter = "";
Joiner.on(delimiter).join(Lists.newArrayList("Your number is ", 47, "!"));
This may be overkill to answer the op's question, but it is good to know about for more complex join operations. This stackoverflow question ranks highly in general google searches in this area, so good to know.
you can use stringbuffer, stringbuilder, and as everyone before me mentioned, "+". I'm not sure how fast "+" is (I think it is the fastest for shorter strings), but for longer I think builder and buffer are about equal (builder is slightly faster because it's not synchronized).
here is an example to read and concatenate 2 string without using 3rd variable:
public class Demo {
public static void main(String args[]) throws Exception {
InputStreamReader r=new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(r);
System.out.println("enter your first string");
String str1 = br.readLine();
System.out.println("enter your second string");
String str2 = br.readLine();
System.out.println("concatenated string is:" + str1 + str2);
}
}
There are multiple ways to do so, but Oracle and IBM say that using +, is a bad practice, because essentially every time you concatenate String, you end up creating additional objects in memory. It will utilize extra space in JVM, and your program may be out of space, or slow down.
Using StringBuilder or StringBuffer is best way to go with it. Please look at Nicolas Fillato's comment above for example related to StringBuffer.
String first = "I eat"; String second = "all the rats.";
System.out.println(first+second);
Using "+" symbol u can concatenate strings.
String a="I";
String b="Love.";
String c="Java.";
System.out.println(a+b+c);

First char to upper case [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
How to upper case every first letter of word in a string?
Most efficient way to make the first character of a String lower case?
I want to convert the first letter of a string to upper case. I am attempting to use replaceFirst() as described in JavaDocs, but I have no idea what is meant by regular expression.
Here is the code I have tried so far:
public static String cap1stChar(String userIdea)
{
String betterIdea, userIdeaUC;
char char1;
userIdeaUC = userIdea.toUpperCase();
char1 = userIdeaUC.charAt(0);
betterIdea = userIdea.replaceFirst(char1);
return betterIdea;
}//end cap1stChar
The compiler error is that the argument lists differ in lengths. I presume that is because the regex is missing, however I don't know what that is exactly.
Regular Expressions (abbreviated "regex" or "reg-ex") is a string that defines a search pattern.
What replaceFirst() does is it uses the regular expression provided in the parameters and replaces the first result from the search with whatever you pass in as the other parameter.
What you want to do is convert the string to an array using the String class' charAt() method, and then use Character.toUpperCase() to change the character to upper case (obviously). Your code would look like this:
char first = Character.toUpperCase(userIdea.charAt(0));
betterIdea = first + userIdea.substring(1);
Or, if you feel comfortable with more complex, one-lined java code:
betterIdea = Character.toUpperCase(userIdea.charAt(0)) + userIdea.substring(1);
Both of these do the same thing, which is converting the first character of userIdea to an upper case character.
Or you can do
s = Character.toUpperCase(s.charAt(0)) + s.substring(1);
public static String cap1stChar(String userIdea)
{
char[] stringArray = userIdea.toCharArray();
stringArray[0] = Character.toUpperCase(stringArray[0]);
return userIdea = new String(stringArray);
}
Comilation error is due arguments are not properly provided, replaceFirst accepts regx as initial arg. [a-z]{1} will match string of simple alpha characters of length 1.
Try this.
betterIdea = userIdea.replaceFirst("[a-z]{1}", userIdea.substring(0,1).toUpperCase())
String toCamelCase(String string) {
StringBuffer sb = new StringBuffer(string);
sb.replace(0, 1, string.substring(0, 1).toUpperCase());
return sb.toString();
}
userIdeaUC = userIdea.substring(0, 1).toUpperCase() + userIdea.length() > 1 ? userIdea.substring(1) : "";
or
userIdeaUC = userIdea.substring(0, 1).toUpperCase();
if(userIdea.length() > 1)
userIdeaUC += userIdea.substring(1);
For completeness, if you wanted to use replaceFirst, try this:
public static String cap1stChar(String userIdea)
{
String betterIdea = userIdea;
if (userIdea.length() > 0)
{
String first = userIdea.substring(0,1);
betterIdea = userIdea.replaceFirst(first, first.toUpperCase());
}
return betterIdea;
}//end cap1stChar

How do I concatenate input in java?

I am trying to concatenate and trying to parse at the same time. I am right now making a excel like program where I can say a1 = "Hello" + "World" and in the cell of A1 have it say HelloWorld. I just need to know how to parse the adding sign and connect those two words. Please tell me if you need more code to understand this, like the runner.
This is my parseInput class :
public class ParseInput {
private static String inputs;
static int col;
private static int row;
private static String operation;
private static Value field;
public static void parseInput(String input){
//splits the input at each regular expression match. \w is used for letters and \d && \D for integers
inputs = input;
Scanner tokens = new Scanner(inputs);
String none0 = tokens.next();
#SuppressWarnings("unused")
String none1 = tokens.next();
operation = tokens.nextLine().substring(1);
String[] holder = new String[2];
String regex = "(?<=[\\w&&\\D])(?=\\d)";
holder = none0.split(regex);
row = Integer.parseInt(holder[1]);
col = 0;
int counter = -1;
char temp = holder[0].charAt(0);
char check = 'a';
while(check <= temp){
if(check == temp){
col = counter +1;
}
counter++;
check = (char) (check + 1);
}
System.out.println(col);
System.out.println(row);
System.out.println(operation);
setField(Value.parseValue(operation));
Spreadsheet.changeCell(row, col, field);
}
public static Value getField() {
return field;
}
public static void setField(Value field) {
ParseInput.field = field;
}
}
This is actually a pretty complicated problem unless you can constrain input to a very small subset of what Excel accepts. If not then you'll probably want to look into something like ANTLR. However, assuming the above input then you'll want to do something like:
Split the string on the equal sign into s1 and s2
Split s2 on the plus sign into s3 and s4.
Trim all the strings, remove the quotes around s3 and s4.
Concatenate s3 and s4 and assign to your datastore indexed by s1.
Depending on how complex your concatenation needs are you can either use string concatenation or a StringBuilder:
result = "" + s3 + s4; // string concatenation
result = new StringBuilder().append(s3).append(s4).toString(); // StringBuilder
Let me know if you have any questions about any of the steps detailed above.
Details on (1) above, assuming input is a1 = "Hello" + "World":
String[] strings = input.split("=");
String s1 = strings[0].trim(); // a1
String s2 = strings[1].trim(); // "Hello" + "World"
strings = s2.split("+");
String s3 = strings[0].trim().replaceAll("^\"", "").replaceAll("\"$", "") // Hello
String s4 = strings[1].trim().replaceAll("^\"", "").replaceAll("\"$", ""); // World
String field = s3 + s4;
String colString = s1.replaceAll("[\\d]", ""); // a
String rowString = s1.replaceAll("[\\D]", ""); // 1
int col = colString.charAt(0) - 'a'; // 0
int row = Integer.parseInt(rowString);
Spreadsheet.changeCell(row, col, field);
I suggest you to implement your custom grammar using a parser generator like JavaCC.
Here you can find a simple tutorial.
I believe this is the better solution because in this way you can handle every expression you need.
Are you sure you want to use all the classes you are using? To parse something like "a=b+c+d.." (assuming you are not trying to validate), easiest and possibly the most efficient way is to use split API in Java lang String
Then join whatever is required using StringBuilder
You need to design and implement a parser and an evaluator. And before that, you need to design the language that your parser/evaluator is going to evaluate.
How to do it.
If your language is really simple, you can get away with parsing it by hand, using something like StringTokenizer to do the tokenization,
Otherwise, you are probably best off learning to use a Java "parser generator" such as JavaCC or ANTLR.
Either way, you need to do some background reading to understand all of the terminology. You could start with Wikipedia and/or the tutorial material from one of the parser generators. Alternatively, there are good textbooks on this topic.
In addition to what Abdullah said, if you really want to save every single ounce of memory you can, you should use the StringBuilder instead of the String concatenation. I believe i read somewhere before that the String concatenation make a new string object for each concatenations while the StringBuilder will add them all to a single String. Shouldn't matter too much though.
In my early life I made an equation evaluator in your style. It cost me huge code and complexity, because of my unawareness about Expression trees. But now with this you will be able to add more capabilities to your parser easily and with native JAVA codes. You will get tons of example of using Expression Trees.

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