String Basics and Memory Representation in DSA Python - Execution Trace

Concept Flow - String Basics and Memory Representation

Create string object

↓

Allocate memory for characters

↓

Store characters sequentially

↓

Add null terminator or length info

↓

Reference string variable points to memory

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Use string in operations (read, slice, concat)

Shows how a string is created in memory, characters stored, and referenced by a variable.

Execution Sample

DSA Python

s = "cat"
print(s)
print(s[1])

Creates a string 'cat', prints it, then prints the character at index 1.

Execution Table

Step	Operation	Memory Allocation	Pointer/Reference	Visual State
1	Create string 'cat'	Allocate 3 chars + length info	s -> address 0x100	[c][a][t] (length=3)
2	Print s	No change	s -> 0x100	Output: cat
3	Access s[1]	No change	s -> 0x100 + offset 1	Output: a
4	End	No change	s -> 0x100	String unchanged

💡 All operations complete, string remains unchanged in memory.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	Final
s	undefined	points to 'cat' at 0x100	points to 'cat' at 0x100	points to 'cat' at 0x100	points to 'cat' at 0x100

Key Moments - 3 Insights

Why does s[1] give 'a' and not 'c'?

Because string indexing starts at 0, s[0] is 'c', s[1] is 'a' as shown in execution_table step 3.

Does printing the string change its memory?

No, printing only reads the string; memory allocation and pointer remain the same as in steps 2 and 4.

Is the string stored as characters or something else?

The string is stored as sequential characters in memory with length info, as shown in step 1's memory allocation.

Visual Quiz - 3 Questions

Look at the execution_table, what is the output when accessing s[1] at step 3?

A "t"

B "c"

C "a"

D Error

💡 Hint

Check the 'Output' column in execution_table row for step 3.

At which step is memory allocated for the string characters?

A Step 1

B Step 2

C Step 3

D Step 4

💡 Hint

Look at the 'Memory Allocation' column in execution_table.

If we change s to "dog", what changes in variable_tracker after step 1?

A s still points to old address with [c][a][t]

B s points to new address with [d][o][g]

C s becomes undefined

D Memory is freed but s points to old address

💡 Hint

Variable tracker shows pointer changes after string creation.

Concept Snapshot

String Basics and Memory Representation:
- Strings are stored as sequential characters in memory.
- A variable points to the memory address of the string.
- Indexing starts at 0 (first char at s[0]).
- Printing or accessing chars does not change string memory.
- Memory includes length info or null terminator for end detection.

Full Transcript

This visual trace shows how a string is created and stored in memory. When we assign s = "cat", memory is allocated for three characters plus length info. The variable s points to this memory. Printing s outputs 'cat' without changing memory. Accessing s[1] returns 'a' because indexing starts at zero. Throughout, the string remains unchanged in memory. This helps understand string basics and how memory references work.