Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Write your own implementation of htonl. Using both your own htonl and (if little-endian hardware is available) the standard library version, run appropriate experiments to determine how much longer it takes to byte-swap integers versus merely copying them.

Short Answer

Expert verified
Implement your own htonl function, then benchmark it against the standard htonl function on little-endian hardware to compare the byte-swapping times.

Step by step solution

01

- Understand htonl

The function htonl stands for 'Host TO Network Long'. It converts an unsigned integer from host byte order to network byte order, which is always big-endian.
02

- Write Your Implementation

Create your own implementation of htonl by writing a function that converts a 32-bit integer from host byte order to network byte order. Example in C:```#include uint32_t my_htonl(uint32_t hostlong) { return ((hostlong & 0xff000000) >> 24) | ((hostlong & 0x00ff0000) >> 8) | ((hostlong & 0x0000ff00) << 8) | ((hostlong & 0x000000ff) << 24);}```
03

- Benchmark Your Function

Set up a benchmarking test to compare the runtime of your function with the standard library’s htonl function on little-endian hardware. For example, use the clock() function in C to measure the time taken:```#include #include int main() { uint32_t value = 0x12345678; clock_t start, end; double cpu_time_used; start = clock(); value = my_htonl(value); end = clock(); cpu_time_used = ((double) (end - start)) / CLOCKS_PER_SEC; printf('my_htonl time: %f', cpu_time_used); start = clock(); value = htonl(value); end = clock(); cpu_time_used = ((double) (end - start)) / CLOCKS_PER_SEC; printf('htonl time: %f', cpu_time_used); return 0;}```
04

- Run Experiments

Execute the benchmarking code on little-endian hardware and record the time taken by each function to perform the conversion.
05

- Analyze Results

Compare the recorded times from the experiments. Determine how much longer it takes to byte-swap integers using your implementation of htonl versus merely copying them using the standard library version.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

endianness
Endianness describes the order in which bytes are stored in memory. There are two main types:
  • Big-endian: Stores the most significant byte (MSB) first.
  • Little-endian: Stores the least significant byte (LSB) first.
A system's endianness impacts how binary data is interpreted. For instance, the number 0x12345678 might be stored differently depending on the system's endianness. Big-endian stores it as 12 34 56 78, while little-endian stores it as 78 56 34 12.
Understanding endianness is crucial for network communication and data interchange between different systems. Network protocols typically use big-endian order, also known as network byte order, to ensure consistency. Functions like htonl (Host TO Network Long) convert integers to network byte order, ensuring that multi-byte data is correctly interpreted regardless of the host system's endianness.
byte swapping
Byte swapping is the process of converting a data value between little-endian and big-endian formats. This is essential when transferring data between systems with different endianness. The process involves rearranging the bytes in the value.
Here's how to swap bytes in a 32-bit integer, using a step-by-step function:
`uint32_t my_htonl(uint32_t hostlong) { return ((hostlong & 0xff000000) >> 24) | ((hostlong & 0x00ff0000) >> 8) | ((hostlong & 0x0000ff00) << 8) | ((hostlong & 0x000000ff) << 24);}`
This rearranges the bytes from little-endian order (32-bit) to big-endian. Without this conversion, the data integrity may be compromised when different-endian systems exchange information.
It's a vital process in network communication and systems interoperability. Functions like htonl and ntohl (Network TO Host Long) automate this conversion, making data handling more reliable.
benchmarking
Benchmarking is measuring the performance of a piece of code or a function to determine its efficiency. In the context of byte swapping, we compare the performance of a custom implementation with the standard library's htonl function. Here's an example benchmarking setup in C:
`#include #include int main() { uint32_t value = 0x12345678; clock_t start, end; double cpu_time_used;
start = clock(); value = my_htonl(value); end = clock(); cpu_time_used = ((double) (end - start)) / CLOCKS_PER_SEC; printf('my_htonl time: %f', cpu_time_used);
start = clock(); value = htonl(value); end = clock(); cpu_time_used = ((double) (end - start)) / CLOCKS_PER_SEC; printf('htonl time: %f', cpu_time_used); return 0;}`
This code measures the time each function takes to perform the byte swap by recording the start and end times and calculating the difference. The results show how much longer custom implementations take compared to the library versions, giving insights into performance differences.
Benchmarking helps identify efficient methods and optimize code, crucial for time-sensitive applications, such as real-time data processing or systems requiring high performance.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The presentation formatting process is sometimes regarded as an autonomous protocol layer, separate from the application. If this is so, why might including data compression in the presentation layer be a bad idea?

Using the programming language of your choice that supports user-defined automatic type conversions, define a type netint and supply conversions that enable assignments and equality comparisons between ints and netints. Can a generalization of this approach solve the problem of network argument marshalling?

Suppose we have a video of two white points moving toward each other at a uniform rate against a black background. We encode it via MPEG. In one I frame the two points are 100 pixels apart; in the next I frame they have merged. The final point of merger happens to lie at the center of a \(16 \times 16\) macroblock. (a) Describe how you might optimally encode the Y component of the intervening B (or P) frames. (b) Now suppose the points are in color, and that the color changes slowly as the points move. Describe what the encoding of the U and V values might look like.

Let \(p \leq 1\) be the fraction of machines in a network that are big-endian; the remaining \(1-p\) fraction are little-endian. Suppose we choose two machines at random and send an int from one to the other. Give the average number of byte-order conversions needed for both big-endian network byte order and receiver-makes-right, for \(p=0.1, p=0.5\), and \(p=0.9\). Hint: The probability that both endpoints are big-endian is \(p^{2} ;\) the probability that the two endpoints use different byte orders is \(2 p(1-p)\).

Suppose you want to implement fast-forward and reverse for MPEG streams. What problems do you run into if you limit your mechanism to displaying I frames only? If you don't, then to display a given frame in the fast-forward sequence, what is the largest number of frames in the original sequence you may have to decode?

See all solutions

Recommended explanations on Computer Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free