In C, mixing char* and int* and unaligned reads: what prints? #include<stdio.h> int main() { int arr[3] = {2, 3, 4}; char p; p = (char)arr; p = (char*)((int*)(p)); printf("%d, ", p); p = (int)(p + 1); printf("%d", p); return 0; }

Introduction / Context:
This question tests your understanding of pointer casts between char and int*, byte-level access, alignment, and the consequences of reading an int from a non–int-aligned address. Many platforms allow the code to compile but may yield implementation-defined or undefined results at runtime when misaligned.

Given Data / Assumptions:

• arr = {2, 3, 4}
• p is a char* pointing to the first byte of arr.
• Ordinary little-endian assumptions explain why the first byte of the first int is 2.
• The code later forms an int* from p + 1, which is typically misaligned.

Concept / Approach:
p reads the very first byte of the array’s memory, which for the integer 2 on little-endian is 0x02 → prints 2. Then p = (int)(p + 1) creates an int* at the following byte boundary. Dereferencing *p reads four bytes starting at a nonaligned address, yielding a “garbage” integer (and possibly a bus error on strict-alignment systems).

Step-by-Step Solution:
Initial: p points to arr as bytes.Print p → 2.Advance by 1 byte and cast to int → potential misalignment.Print *p → implementation-defined “garbage” value.

Verification / Alternative check:
On a platform that traps misaligned access, the program may fault. On tolerant CPUs, the value varies by endianness and representation.

Why Other Options Are Wrong:
(a) assumes a clean aligned int read. (b) suggests a forced zero which is unlikely. (d) prints zeros without basis.

Common Pitfalls:
Forgetting alignment rules; assuming bytes map to integers uniformly across architectures.

Final Answer:
2, Garbage value