String creation in Java - Step-by-Step Execution

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Concept Flow - String creation

Start

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Declare String variable

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Assign value using literal or new

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String object created in memory

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Variable holds reference to String

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Use String in program

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End

This flow shows how a String variable is declared, assigned a value, and holds a reference to a String object in memory.

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Execution Sample

Java

String s1 = "Hello";
String s2 = new String("World");
System.out.println(s1 + " " + s2);

Creates two strings in different ways and prints them with a space.

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Execution Table

Step	Action	Variable	Value/Reference	Output
1	Declare s1 and assign literal "Hello"	s1	Reference to "Hello" in string pool
2	Declare s2 and assign new String("World")	s2	Reference to new String object "World"
3	Print s1 + " " + s2			Hello World
4	End of program

💡 Program ends after printing the combined string.

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Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	Final
s1	undefined	Reference to "Hello"	Reference to "Hello"	Reference to "Hello"	Reference to "Hello"
s2	undefined	undefined	Reference to new String "World"	Reference to new String "World"	Reference to new String "World"

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Key Moments - 2 Insights

▶Why does s1 point to the string pool but s2 points to a new object?

▶Does s1 == s2 return true?

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Visual Quiz - 3 Questions

Look at the execution table, what does variable s1 reference after step 1?

A A string literal in the string pool

B A new String object in heap

C null

D An integer value

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Concept Snapshot

String creation in Java:
- Use String s = "text"; to create from literal (stored in string pool).
- Use String s = new String("text"); to create a new object.
- Variables hold references to String objects.
- == checks references, .equals() checks content.
- String literals reuse objects for memory efficiency.

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Full Transcript

This example shows how Java creates strings. First, s1 is declared and assigned a string literal "Hello", which Java stores in a special area called the string pool. Then s2 is declared and assigned a new String object with the text "World". This creates a new object in memory, different from the string pool. When printing s1 and s2 combined with a space, the output is "Hello World". The variable tracker shows s1 and s2 holding references to different objects. Beginners often confuse string literals with new String objects and the difference in memory. Also, the == operator compares references, so s1 == s2 is false here. Changing s2 to a literal would make it reference the string pool like s1. This helps save memory and allows string reuse.