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Converting and storing integers to ASCII Characters [duplicate]

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);

Is String.trim() faster than String.replace()?

Let's say there has a string like " world ". This String only has the blank at front and end. Is the trim() faster than replace()?
I used the replace once and my mentor said don't use it since the trim() probably faster.
If not, what's the advantage of trim() than replace()?

If we look at the source code for the methods:
replace():
public String replace(CharSequence target, CharSequence replacement) {
String tgtStr = target.toString();
String replStr = replacement.toString();
int j = indexOf(tgtStr);
if (j < 0) {
return this;
}
int tgtLen = tgtStr.length();
int tgtLen1 = Math.max(tgtLen, 1);
int thisLen = length();
int newLenHint = thisLen - tgtLen + replStr.length();
if (newLenHint < 0) {
throw new OutOfMemoryError();
}
StringBuilder sb = new StringBuilder(newLenHint);
int i = 0;
do {
sb.append(this, i, j).append(replStr);
i = j + tgtLen;
} while (j < thisLen && (j = indexOf(tgtStr, j + tgtLen1)) > 0);
return sb.append(this, i, thisLen).toString()
}
Vs trim():
public String trim() {
int len = value.length;
int st = 0;
char[] val = value; /* avoid getfield opcode */
while ((st < len) && (val[st] <= ' ')) {
st++;
}
while ((st < len) && (val[len - 1] <= ' ')) {
len--;
}
return ((st > 0) || (len < value.length)) ? substring(st, len) : this;
}
As you can see replace() calls multiple other methods and iterates throughout the entire String, while trim() simply iterates over the beginning and ending of the String until the character isn't a white space. So in the single respect of trying to only remove white space before and after a word, trim() is more efficient.
We can run some benchmarks on this:
public static void main(String[] args) {
long testStartTime = System.nanoTime();;
trimTest();
long trimTestTime = System.nanoTime() - testStartTime;
testStartTime = System.nanoTime();
replaceTest();
long replaceTime = System.nanoTime() - testStartTime;
System.out.println("Time for trim(): " + trimTestTime);
System.out.println("Time for replace(): " + replaceTime);
}
public static void trimTest() {
for(int i = 0; i < 1000000; i ++) {
new String(" string ").trim();
}
}
public static void replaceTest() {
for(int i = 0; i < 1000000; i ++) {
new String(" string ").replace(" ", "");
}
}
Output:
Time for trim(): 53303903
Time for replace(): 485536597
//432,232,694 difference

Assuming that the people writing the Java library code are doing a good job1, you can assume that a special purpose method (like trim()) will be as fast, and probably faster than a general purpose method (like replace(...)) doing the same thing.
Two reasons:
If the special purpose method is slower, its implementation can be rewritten as equivalent calls to the general purpose one, making the performance equivalent in most cases. A competent programmer will do this because it reduces maintenance costs.
In the special purpose method, it is likely that there will be optimizations that can be made that don't apply in the general-purpose case.
In this case we know that trim() only needs to look at the start and end of the string ... whereas replace(...) needs to look at all of the characters in the string. (We can infer this from the description of what the respective methods do.)
If we assume "competence" then we can infer that the developers will have done the analysis and not implemented trim() sub-optimally2; i.e. they won't code trim() to examine all characters.
There is another reason to use the special purpose method over the general purpose. It makes your code simpler, easier to read, and easier to inspect for correctness. This may well be more important than performance.
This clearly applies in the case of trim() versus replace(...).
1 - We can in this case. There are lots of eyes looking at this code, and lots of people who will complain loudly about egregious performance issues.
2 - Unfortunately, it is not always as straightforward as this. A library method needs to be optimized for "typical" behavior, but it also needs to avoid pathological performance in edge-cases. It is not always possible to achieve both things.

trim() is definitely faster to type, yes. It doesn't take any parameters.
It is also much faster to understand what you where trying to do. You were trying to trim the string, rather than replacing all the spaces it contains with the empty string, knowing from other context that there is only space at the beginning and the end of the string.
Indeed much faster no matter how you look at it. Don't complicate the life of the persons who're trying to read your code. Most of the time, it will be you months later, or at least someone you don't hate.

Trim will prune the outter characters until they are non white space. I believe they trim space, tab, and new lines.
Replace will scan the entire string (so, it could be a sentense) and would replace inner " " with "", essentially compressing them together.
They have different use cases though, obviously 1 is to clean up user input where the other is to update a string where matches are found with something else.
That being said, run times: Replace will run in N time, as it will look for all matching characters. Trim will run in O(N), but most likely a just a few characters off of each end.
The idea behind trim i think came around from people would would type and input things but accidentally press space before submitting their forms, essentially trying to save the field "Foo " instead of "Foo"

s.trim() shortens a String s. This means no characters has to be moved from an index to another. It starts at the first character (s.toCharArray()[0]) of the String and shortens the String character by character until the first non-whitespace character occurs. It works the same way to shorten the String at the end. So it compresses the String. If a String has no leading and trailing whitespace trim will be ready after checking the first and the last character.
In case of " world ".trim() two steps are needed: one to remove the first leading whitespace as it is on the first index and the the second to remove the last whitespace as it is on the last index.
" world ".replace(" ", "") will need at least n = " world ".length() steps. It has to check every character if it has to be replaced. But if we take into account that the implementation of String.replace(...) needs to compile a Pattern, build a Matcher and then to replace all the matched regions it's seems far complex comparing to shorten a String.
We also have to consider that " world ".replace(" ", "") does not replace whitespaces but only the String " ". Since String replace(CharSequence target, CharSequence replacement) compiles the target using Pattern.LITERAL we cannot use the character class \s. To be more accurate we would have to compare " world ".trim() to " world ".replaceAll("\\s", ""). It is still not the same because a whitespace in String trim() is defined as c <= ' ' for each c in s.toCharArray().
Summarizing: String.trim() should be faster - especially for long strings
The description how the methods work is based on the implementation of String in Java 8. But implementations can change.
But the question should be: What do you intent to do with the string? Do you want to trim it or to replace some characters? According to it use the corresponding method.

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);

Strings and Loops Practice Test

I'm slightly confused on this test question. I made a chart of the values of i , j, and the string. I got "nbearig", but my runtime is printing out numbers. I'm not sure where I went wrong. ++i , --j means that they were incre/decremented before the code after the for loop right?
public class AlGore {
public static void main(String[] args) {
String mystery = "mnerigpaba";
String solved = "";
int len = mystery.length();
for (int i = 0, j = len - 1; i < len/2; ++i, --j) {
solved += mystery.charAt(i) + mystery.charAt(j);
}
System.out.println(solved);
}
}

I'm not sure where I went wrong. ++i , --j means that they were incre/decremented before the code after the for loop right?
1) They were preincremented / predecremented respectively.
2) It happened after each execution of the loop body.
my compiler is printing out numbers.
No it isn't. The compiler is compiling your code!!! The JVM is printing numbers ... when you run the code.
To understand the reason why, take a careful look at this:
solved += mystery.charAt(i) + mystery.charAt(j);
This is equivalent to
solved = solved + ( mystery.charAt(i) + mystery.charAt(j) );
Now the expression in brackets performs a numeric addition of a character to a character. According to the rules of Java expressions, that gives an int value. So the entire expression becomes:
solved = String.concat(
solved,
Integer.toString(mystery.charAt(i) + mystery.charAt(j));
I thought that the charAt(i) function will return a string?
No. It returns a char ... just like the method name "charAt" implies. String and char are fundamentally different types.
Comment: That is a good exam question, it tests how well you understand loops, and how well you understand Java expression semantics.

You are performing integer math (because char is an integral type),
// solved += mystery.charAt(i) + mystery.charAt(j);
solved += Character.toString(mystery.charAt(i))
+ Character.toString(mystery.charAt(j));
That way you are performing String concatenation.

mystery.charAt(i) + mystery.charAt(j); will add the numeric values of those two characters. You can force string concatenation by adding "" + before:
solved += "" + mystery.charAt(i) + mystery.charAt(j);

Printing the alphabet using for loop. Why must I explicitly convert character [] to string?

I am trying to print the letters of the alphabet in caps. So I wrote this in a for loop:
System.out.print(Character.toChars(i));
//where i starts at 65 and ends at 90
This works fine and prints the letters but In my code I wanted to put a space between the letters to make it look nicer. So i did this:
System.out.print(Character.toChars(i) + " ")
Why does it print the memory address of the characters instead of the letter?
The solution I came up with was to explicitly convert the char to a new String object:
String character = new String(Character.toChars(i));
System.out.print (character + " ");
but I'm not quite sure why I can't just write "Character.toChars(i)"
In the first one Does the method(Character.toChars()) point to the address of the character and System.out.print is smart enough to print the value at that address? i.e the corresponding letter?

System.out.print(Character.toChars(i)) calls PrintStream.print(char[]), an overload that handles char[] specially.
Character.toChars(i) + " " is really equivalent to Character.toChars(i).toString() + " "; calling toString() on an array type results in a string representation of its address (this behaviour is directly inherited from Object).
A simpler solution for your particular case may be this:
System.out.println((char)i + " ");

The Character.toChars method returns char[], which will be represented as [C#<hex hashcode> in String form.
You don't need to use the toChars method (or do any casting at all):
for (char c = 'A'; c <= 'Z'; c++) {
System.out.print(c + " ");
}

You use string concatenation, with one side being an array of chars and the other a string and according to the Java language specification, then as the char array is not a primitive type, but a reference value (aka an object), its toString method is called. And as there is no specific method implemented for arrays, they inherit the method implementation from java.lang.Object, which prints the address.
On the other hand, System.out.print(Character.toChars(i)) calls a specific implementation of print for character arrays, see the documentation of PrintStream.