var i = 0, count = 1000, a;
var map = new Map();
for (i = 0; i < count; i++) {
if (Math.random() > 0.5) {
map.set(i, i * i);
}
}
for (i = 0; i < count; i++) {
a = map.has(i);
}
for (i = 0; i < count; i++) {
a = map.get(i);
}
--enable-precise-memory-info
flag.
Test case name | Result |
---|---|
Map 1 | |
Map 2 |
Test name | Executions per second |
---|---|
Map 1 | 5669.4 Ops/sec |
Map 2 | 5651.5 Ops/sec |
Let's dive into the world of JavaScript microbenchmarks.
Benchmark Overview
The provided JSON represents two individual test cases for measuring the performance difference between Map.get()
and Map.has()
. The benchmark is designed to compare these two methods in terms of execution speed on a map data structure.
Options Compared
Two options are compared:
Map.get(i)
: This method retrieves the value associated with a given key (i
) from the map.Map.has(i)
: This method checks if a key is present in the map without retrieving its value.Pros and Cons of Each Approach
Map.get(i)
: This approach involves retrieving the value associated with each key. If the key exists, it returns the corresponding value; otherwise, it returns undefined
.
undefined
).Map.has(i)
: This approach involves checking if a key exists without retrieving its value. It returns a boolean indicating whether the key is present.
Performance Considerations: Since Map.has(i)
involves checking for existence without retrieving values, it generally outperforms Map.get(i)
when the key is not present. However, if the key exists and its value needs to be retrieved, Map.get(i)
might provide better performance due to direct access.
Library Usage
There is no explicit library usage mentioned in the provided JSON. Both Map.get()
and Map.has()
are native JavaScript methods.
Special JS Feature/Syntax
No special JavaScript features or syntax are used in this benchmark.
Alternative Approaches
Some alternative approaches could be explored:
Set
data structure might provide better performance. However, if you need to store key-value pairs, a Map
is still the most suitable choice.Map.has(i)
or Map.get(i)
for repeated executions on the same iteration index.By understanding these nuances, developers can make informed decisions about optimizing their code's performance in various scenarios.