String Basics and Memory Representation in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of String Pointer Arithmetic

What is the output of the following C code snippet?

DSA C

char *str = "hello";
printf("%c", *(str + 1));

A e

B h

C o

D l

💡 Hint

Remember that pointer arithmetic moves to the next character in the string.

The pointer 'str' points to the first character 'h'. Adding 1 moves to the second character 'e'. Dereferencing gives 'e'.

❓ Predict Output

intermediate

String Length Calculation Using Pointer

What is the value of 'length' after running this code?

DSA C

char *s = "abcde";
int length = 0;
while (*(s + length) != '\0') {
    length++;
}

A 4

B 5

C 6

D 0

💡 Hint

Count characters until the null terminator '\0'.

The loop increments length until it reaches the null character, counting all 5 characters.

🧠 Conceptual

advanced

Understanding String Memory Layout

Which statement correctly describes how the string "data" is stored in memory in C?

A Characters are stored backwards in memory ending with '\0'.

B Characters are stored in random memory locations with a pointer array pointing to each.

C The string is stored as a single integer representing all characters.

D Each character is stored in consecutive memory locations followed by a null terminator '\0'.

💡 Hint

Think about how arrays and strings are stored in C.

In C, strings are arrays of characters stored consecutively in memory, ending with a null terminator.

🔧 Debug

advanced

Identify the Error in String Modification

What error occurs when running this code snippet?

DSA C

char *str = "hello";
str[0] = 'H';

A Segmentation fault (attempt to modify read-only memory)

B No error, output is 'Hello'

C Compilation error: cannot assign to str[0]

D Runtime error: null pointer dereference

💡 Hint

String literals are stored in read-only memory.

String literals are stored in read-only memory; modifying them causes a segmentation fault.

🚀 Application

expert

Result of Pointer Increment on String Array

Given the code below, what is printed?

DSA C

char arr[] = "abc";
char *p = arr;
printf("%c", *++p);

A \0

B a

C b

D c

💡 Hint

The operator ++p increments the pointer before dereferencing.

The pointer p is incremented to point to 'b', then dereferenced to print 'b'.