### Problem 2:Assume we have the following code:```cuint32_t arr32[] = {0x11220102, 0x33440304, 0x55660506};uint16_t *ptr16 = (uint16_t *)arr32;uint8_t *ptr8 = (uint8_t *)ptr32++;printf("0x%x\n", *(uint16_t *)ptr32);printf("0x%x\n", (uint8_t)*(ptr16+1));```What are the values printed out from the above code?---### Explanation:This code involves pointer manipulation and type casting in C. The array `arr32` is of type `uint32_t`, which means each element is 4 bytes. The `ptr16` and `ptr8` pointers cast this array to `uint16_t` and `uint8_t` types, respectively.1. **`ptr32++`**: The original pointer to a `uint32_t` is incremented, moving to the next 4-byte block in the array.2. **First `printf`**: It prints the value at the new location to which `ptr32` points, cast as a `uint16_t` (since the pointer is incremented, it points to the second element, `0x33440304`).3. **Second `printf`**: It prints the second byte (using the first element of `arr32`), accessed via the `ptr16` pointer.This test assesses understanding of pointer arithmetic, data representation, and casting in C.### Output:The handwritten notes on the side suggest the expected output values:- `0x102`- `0x10`

uint32_t arr32[] = {0x11220102, 0x33440304, 0x55660506};an array arr32 containing three 32-bit…

Answered: Prob 2 uint32_t uint16_t printf("C…

Computer Networking: A Top-Down Approach (7th Edition)

7th Edition

ISBN:9780133594140

Author:James Kurose, Keith Ross

Publisher:James Kurose, Keith Ross

Chapter1: Computer Networks And The Internet

Section: Chapter Questions

Problem R1RQ: What is the difference between a host and an end system? List several different types of end...

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Related questions

Question

### Problem 2:

Assume we have the following code:

```c
uint32_t arr32[] = {0x11220102, 0x33440304, 0x55660506};
uint16_t *ptr16 = (uint16_t *)arr32;
uint8_t *ptr8 = (uint8_t *)ptr32++;
printf("0x%x\n", *(uint16_t *)ptr32);
printf("0x%x\n", (uint8_t)*(ptr16+1));
```

What are the values printed out from the above code?

---

### Explanation:

This code involves pointer manipulation and type casting in C. The array `arr32` is of type `uint32_t`, which means each element is 4 bytes. The `ptr16` and `ptr8` pointers cast this array to `uint16_t` and `uint8_t` types, respectively.

1. **`ptr32++`**: The original pointer to a `uint32_t` is incremented, moving to the next 4-byte block in the array.
2. **First `printf`**: It prints the value at the new location to which `ptr32` points, cast as a `uint16_t` (since the pointer is incremented, it points to the second element, `0x33440304`).
3. **Second `printf`**: It prints the second byte (using the first element of `arr32`), accessed via the `ptr16` pointer.

This test assesses understanding of pointer arithmetic, data representation, and casting in C.

### Output:

The handwritten notes on the side suggest the expected output values:
- `0x102`
- `0x10`

Expert Solution

Step 1: Array Initialization and Pointer Creation:

uint32_t arr32[] = {0x11220102, 0x33440304, 0x55660506};

an array arr32 containing three 32-bit unsigned integers (uint32_t). The array is initialized with these hexadecimal values: 0x11220102, 0x33440304, and 0x55660506.

uint32_t *ptr32 = arr32;

Create a pointer ptr32 of type uint32_t* and set it to point to the first element of the arr32 array. This means ptr32 points to the first 32-bit value, 0x11220102.

uint16_t *ptr16 = (uint16_t *)arr32;

Create another pointer ptr16 of type uint16_t*. You initialize it by casting the arr32 array to a pointer of uint16_t. This means ptr16 points to the same memory location as arr32, but it interprets the data as 16-bit values.