Number of objects created during string concatenation

Question

Can someone please tell me how many objects will be created on executing the System.out.println statement in the below code

int i=0;
int j=1;
System.out.print("i value is "+ i + "j value is "+j);

Answer 1

If you really want to know what's going on, why not look at the bytecode?

I wrapped your code in a main function, compiled it and then disassembled it with javap -c Test.class . Here's the output (using a Oracle Java 7 compiler):

Compiled from "Test.java"
class Test {
  Test();
    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_0
       1: istore_1
       2: iconst_1
       3: istore_2
       4: getstatic     #2                  // Field java/lang/System.out:Ljava/io/PrintStream;
       7: new           #3                  // class java/lang/StringBuilder
      10: dup
      11: invokespecial #4                  // Method java/lang/StringBuilder."<init>":()V
      14: ldc           #5                  // String i value is
      16: invokevirtual #6                  // Method java/lang/StringBuilder.append:(Ljava/lang/String;)Ljava/lang/StringBuilder;
      19: iload_1
      20: invokevirtual #7                  // Method java/lang/StringBuilder.append:(I)Ljava/lang/StringBuilder;
      23: ldc           #8                  // String j value is
      25: invokevirtual #6                  // Method java/lang/StringBuilder.append:(Ljava/lang/String;)Ljava/lang/StringBuilder;
      28: iload_2
      29: invokevirtual #7                  // Method java/lang/StringBuilder.append:(I)Ljava/lang/StringBuilder;
      32: invokevirtual #9                  // Method java/lang/StringBuilder.toString:()Ljava/lang/String;
      35: invokevirtual #10                 // Method java/io/PrintStream.print:(Ljava/lang/String;)V
      38: return
}

The only object that gets allocated in this method is the StringBuilder (by the new instruction at position 7). However, the other methods that are invoke d might allocated something themselves, and I have a very strong suspicion that StringBuilder.toString will allocate a String object.

Answer 2

这将创建一个StringBuilder对象（以及此对象在内部使用的任何对象），添加值，最后StringBuilder将创建一个带有结果的String对象。

Answer 3

The code

int i=0;
int j=1;
System.out.print("i value is "+ i + "j value is "+j);

Creates 3 objects.

My Reason:

The basic data types in Java are not objects and does not inherit from Object. so int i=0; and int j=1; int i=0; and int j=1; does not make an object.

Now System.out.print("i value is "+ i + "j value is "+j); which contains String which are immutable, and the operations on string are costly.We can split the operations as this.

("i value is ").concat(i)  // creates one object let say obj1
obj1.concat("j value is ")   //creates another let say obj2
obj2.concat(j)            // creates the final string let say obj3;

In an example string operation str1.concat(str2) is done by using two String objects and it creates the third one and change the reference making an illusion that its actually the first string object ie str1. Thus the str1 will be having a new String which contains the value of the old str1 and the str2 concatenated.

This is what i believe with my limited knowledge. Correct me if i am wrong.

Answer 4

只有1 ，String对象被连​​接起来。

Answer 5

If it is indeed implementation dependent, then if it is evaluated to:

StringBuilder sb = new StringBuilder("i value is ");
sb.append(i);
sb.append(j);
String newStr = sb.toString();

There will be 2 objects.

Answer 6

Only 1 ie for printing the String .

Thumb rule - Whenever concatenation of strings are done , a new object is created.

Answer 7

The code is converted, by the compiler, to something like this:

int i=0;
int j=1;
StringBuilder temp = new StringBuilder(); // creates a StringBuilder
temp.append("i value is "); // creates or re-uses a String
temp.append(i); // might create a String
temp.append("j value is"); // creates or re-uses a String
temp.append(j); // might create a String
String temp2 = temp.toString(); // creates a String
System.out.print(temp2);

It depends on whether you count the "i value is " and "j value is " strings, which are created once and then re-used.

If you do count them, then at least 4, otherwise at least 2.

Actually, each String has its own char[] that actually stores the string. So that's 7 or 3, instead of 4 or 2.

StringBuilder has a char[] as well and might have to create new char[]'s as you add more data to it. String.valueOf or System.out.print might also create objects behind your back and there's no way you can know about them without external tools. Hence "at least".

Answer 8

String is immutable means that you cannot change the object itself.
While performing concatenation every time new string objects are created.
So in above example

1. int i
2. int j
3. "i value is "
4. "i value is "+ i
5. "i value is "+ i + "j value is "
6. "i value is "+ i + "j value is "+j

Total six objects are created in above example.

Answer 9

5 objects

1. "i value is ".A String object will be formed.
2. "i value is " +i .This concatenation operation will form 2nd Object.
3. "j value is " will form third object.
4. "i value is "+i + "j value is ".4th object formed because of concatenation.
5. "i value is "+ i + "j value is " + j .This last concatenation will form 5th object.

Answer 10

You should consider 2 points here:

1. When you say String is immutable, it will create a new object if you try to change the value.
2. String object will be created when you write code using Literal(String s = "Pool") way or using new keyword (String s = new String("Heap and Pool"), here new keyword refers heap, the Literal "Heap and Pool" refers String constant pool).

In this case the above answer is correct. Five String objects will be created.