There are multiple ways:

• String.valueOf(number) (my preference)
• "" + number (I don't know how the compiler handles it, perhaps it is as efficient as the above)
• Integer.toString(number)

Normal ways would be Integer.toString(i) or String.valueOf(i).

The concatenation will work, but it is unconventional and could be a bad smell as it suggests the author doesn't know about the two methods above (what else might they not know?).

Java has special support for the + operator when used with strings (see the documentation) which translates the code you posted into:

StringBuilder sb = new StringBuilder();
sb.append("");
sb.append(i);
String strI = sb.toString();

at compile-time. It's slightly less efficient (sb.append() ends up calling Integer.getChars(), which is what Integer.toString() would've done anyway), but it works.

To answer Grodriguez's comment: ** No, the compiler doesn't optimise out the empty string in this case - look:

simon@lucifer:~$ cat TestClass.java
public class TestClass {
  public static void main(String[] args) {
    int i = 5;
    String strI = "" + i;
  }
}
simon@lucifer:~$ javac TestClass.java && javap -c TestClass
Compiled from "TestClass.java"
public class TestClass extends java.lang.Object{
public TestClass();
  Code:
   0:    aload_0
   1:    invokespecial    #1; //Method java/lang/Object."<init>":()V
   4:    return

public static void main(java.lang.String[]);
  Code:
   0:    iconst_5
   1:    istore_1

Initialise the StringBuilder:

   2:    new    #2; //class java/lang/StringBuilder
   5:    dup
   6:    invokespecial    #3; //Method java/lang/StringBuilder."<init>":()V

Append the empty string:

   9:    ldc    #4; //String
   11:    invokevirtual    #5; //Method java/lang/StringBuilder.append:
(Ljava/lang/String;)Ljava/lang/StringBuilder;

Append the integer:

   14:    iload_1
   15:    invokevirtual    #6; //Method java/lang/StringBuilder.append:
(I)Ljava/lang/StringBuilder;

Extract the final string:

   18:    invokevirtual    #7; //Method java/lang/StringBuilder.toString:
()Ljava/lang/String;
   21:    astore_2
   22:    return
}

There's a proposal and ongoing work to change this behaviour, targetted for JDK 9.

You can use sprintf to do it, or maybe snprintf if you have it:

char str[ENOUGH];
sprintf(str, "%d", 42);

Where the number of characters (plus terminating char) in the str can be calculated using:

(int)((ceil(log10(num))+1)*sizeof(char))