String Traversal and Character Access in DSA C - Execution Trace

Concept Flow - String Traversal and Character Access

Start at index = 0

↓

Check: index < string length?

No→EXIT

Yes↓

Access character at index

↓

Process character (e.g., print)

↓

Increment index by 1

↩Back to Check

This flow shows how we start from the first character of a string, check if we are still inside the string, access and process the character, then move to the next one until we reach the end.

Execution Sample

DSA C

char str[] = "hello";
int i = 0;
while (str[i] != '\0') {
    printf("%c", str[i]);
    i++;
}

This code prints each character of the string "hello" one by one by accessing each character using its index.

Execution Table

Step	i (index)	Condition (str[i] != '\0')	Character Accessed	Action	Output
1	0	h != '\0' (True)	'h'	Print 'h', i++	h
2	1	e != '\0' (True)	'e'	Print 'e', i++	he
3	2	l != '\0' (True)	'l'	Print 'l', i++	hel
4	3	l != '\0' (True)	'l'	Print 'l', i++	hell
5	4	o != '\0' (True)	'o'	Print 'o', i++	hello
6	5	'\0' != '\0' (False)	N/A	Exit loop	hello

💡 At step 6, str[5] is '\0', so condition is false and loop ends.

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	Final
i	0	1	2	3	4	5	5

Key Moments - 3 Insights

Why does the loop stop when str[i] is '\0'?

The '\0' character marks the end of the string. When str[i] equals '\0', the condition str[i] != '\0' becomes false, so the loop stops as shown in execution_table step 6.

Why do we start with i = 0 instead of 1?

String indices start at 0 in C. So str[0] is the first character. This is why the loop starts with i = 0, as shown in execution_table step 1.

What happens if we forget to increment i inside the loop?

If i is not incremented, the loop will keep accessing the same character forever, causing an infinite loop. The execution_table shows i increasing each step to avoid this.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of i at step 4?

A 4

B 3

C 2

D 5

💡 Hint

Check the 'i (index)' column in execution_table row for step 4.

At which step does the loop condition become false?

A Step 5

B Step 4

C Step 6

D Step 7

💡 Hint

Look at the 'Condition' column where it shows False in execution_table.

If the string was "hi" instead of "hello", how many times would the loop run?

A 2 times

B 3 times

C 5 times

D 1 time

💡 Hint

Check how many characters before '\0' in the string and compare with execution_table steps.

Concept Snapshot

String Traversal and Character Access in C:
- Use an index starting at 0 to access characters.
- Loop until you reach the '\0' end character.
- Access characters with str[i].
- Increment i each loop to move forward.
- Stop when str[i] == '\0'.

Full Transcript

This lesson shows how to go through each character of a string in C using an index. We start at index 0, check if the character is not the end marker '\0', print the character, then move to the next index. We repeat until we reach '\0', which means the string ended. The code example prints 'hello' one letter at a time. The execution table tracks the index, condition, character accessed, and output at each step. Key points include why the loop stops at '\0', why indexing starts at 0, and the importance of incrementing the index to avoid infinite loops.