Benchmark: Math.pow(x,0.5) vs x ** 0.5

Script Preparation code:

var numbers = Array.from(Array(10000), (_,x) => (Math.random()*x));

Tests:

sqrt with Math.pow
numbers.forEach(x => Math.pow(x,0.5));
x ** 0.5
numbers.forEach(x => x ** 0.5);

Rendered benchmark preparation results:

Suite status: <idle, ready to run>

Previous results

Test case name	Result
sqrt with Math.pow
x ** 0.5

Fastest: N/A

Slowest: N/A

Latest run results:

Run details: (Test run date: one year ago)

User agent: Mozilla/5.0 (Linux; Android 13) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/113.0.0.0 Mobile Safari/537.36

Browser/OS: Chrome Mobile 113 on Android

View result in a separate tab

Test name	Executions per second
sqrt with Math.pow	281.8 Ops/sec
x ** 0.5	35428.0 Ops/sec

Autogenerated LLM Summary (model llama3.2:3b, generated 2 months ago):

LLMs can make mistakes. Check important info.

I'll break down the explanation for you.

The provided JSON represents a benchmark test case on MeasureThat.net, which compares the performance of two approaches to calculate the square root of numbers: `Math.pow(x, 0.5)` and `x ** 0.5`.

**Options Compared:**

1. **`Math.pow(x, 0.5)`**: This approach uses the `pow()` method to raise a number to a fractional exponent. In this case, it calculates `(x ^ 0.5)`.
2. **`x ** 0.5`**: This approach uses the exponentiation operator (`**`) to raise `x` to the power of 0.5.

**Pros and Cons:**

*   **`Math.pow(x, 0.5)`**:
    *   Pros:
        *   More explicit and readable syntax.
        *   Works with non-numeric types (e.g., strings) without raising an error.
    *   Cons:
        *   Slower than the exponentiation operator (`**`) due to the overhead of the `pow()` method.
*   **`x ** 0.5`**:
    *   Pros:
        *   Faster than using `Math.pow(x, 0.5)` due to the optimized implementation in the JavaScript engine.
        *   More concise and expressive syntax.
    *   Cons:
        *   Less readable syntax for developers unfamiliar with exponentiation operators.

**Library/Language Features:**

*   The test case uses an array of random numbers generated using `Array.from(Array(10000), (_, x) => (Math.random() * x))`. This function creates a new array with 10,000 elements, each containing a random number between 0 and the current index value (`x`).
*   No specific library or syntax is used in this test case.

**Special JS Features/Syntax:**

*   None mentioned in the provided explanation.

**Other Alternatives:**

If you want to calculate the square root of numbers, other approaches could include:

*   **Using `Math.sqrt()`**: This method directly calculates the square root of a number.
*   **Implementing binary search algorithm**: For large datasets, this can be an efficient way to find approximate square roots.

However, for small to medium-sized datasets like in this benchmark test case, using the exponentiation operator (`**`) or `Math.pow()` is likely sufficient and efficient enough.

Related benchmarks:

Math.pow(x,0.5) vs x ** 0.5 (version: 0)

Comparing performance of: sqrt with Math.pow vs x ** 0.5

Created: one year ago by: Guest

Jump to the latest result

sqrt with Math.pow

x ** 0.5

Suite status: <idle, ready to run>

Experimental features:

Fastest: N/A

Slowest: N/A

Autogenerated LLM Summary (model llama3.2:3b, generated 2 months ago):