hashmapTest | HashMap performance tests from java-performance.info | Performance Testing library
kandi X-RAY | hashmapTest Summary
kandi X-RAY | hashmapTest Summary
HashMap performance tests from java-performance.info.
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Top functions reviewed by kandi - BETA
- Adds an int to the map .
- Remove a key .
- Shift keys with given position .
- Rehashes the keys .
- Gets the read index .
- Insert null .
- Remove null key .
- Returns the next power of two .
- Compute an array size .
hashmapTest Key Features
hashmapTest Examples and Code Snippets
Community Discussions
Trending Discussions on hashmapTest
QUESTION
Recently on the interview I was asked the question:
What will happen if we have two equal objects and we put them as values using the same key? Will the first value be replaced or does hashmap uses
equals()
for values in order to determine whether or not the element already exists?
I answered that if element already present in the bucket than it won't be replaced nor duplicate element will be added.
However, I tried to code this and I see that it's not true. The old object will be replaced.
I have User
entity(randomId
is used to determine which object is currently in the HashMap):
ANSWER
Answered 2018-Feb-03 at 11:00The documentation for put
says:
Associates the specified value with the specified key in this map (optional operation). If the map previously contained a mapping for the key, the old value is replaced by the specified value. (A map m is said to contain a mapping for a key k if and only if m.containsKey(k) would return true.)
(my emphasis)
Note that it says nothing about checking whether the value equals anything, nor should it. The coder has expressly called put
, saying "Put this value in the map under this key." It's not for the map to second-guess the programmer writing that code. It would also be unnecessary overhead.
There can be all kinds of reasons a coder wants to have a specific object, not just an equivalent one, in the map. Canonical caching (à la String#intern
) is probably the first that comes to mind.
QUESTION
I am using BlueJ and testing the HashMap class to see how it works. Below is the code I used to test the class. There is an error that is thrown in line 23, during the first attempt to call the fillMyMap()
method in the constructor.
I tried deleting the call to fillMyMap()
in the constructor. The HashMapTester object is instantiated, but the same NullPointerException
is thrown when I call that method explicitly.
I tried rewriting the myMap
variable declaration, but using a different syntax results in a failure to compile.
I've tested other HashMap
code (e.g., from Objects First with BlueJ), and that code runs fine, so there is not a Library, class, or package problem.
I tried changing the variables, thinking I accidentally hit a reserved word. The same result. What is wrong with this code?
...ANSWER
Answered 2017-Jul-19 at 19:35You're creating a local variable in your constructor. You're not initializing the instance variable.
QUESTION
I am asking this question with respect to java version till 1.7 only. I am using reflection to find out current capacity of HashMap. In below program am putting 12 unique person into a single bucket of HashMap (using same hashcode) . Then i am putting 13th unique person on same or different bucket(using same or different hashcodes). In both the cases after adding this 13th element, HashMap resizes to 32 buckets. i understand that due to load factor .75 and initial capacity 16 HashMap resizes to it's double with 13th element. But there are still empty buckets available and only 2 buckets are used for these 13th element.
My questions are:
1) Is my understanding correct. Am i not making any mistake. Is this the expected behavior of HashMap .
2) If all this is correct then even though there are 12 or 11 free buckets why the need to double the HashMap with 13th element in this case. Isn't it extra overhead or costly to resize the HashMap. What is the need to double the HashMap in this case While 13th can be put in any avalable bucket according to hashcode.
...ANSWER
Answered 2017-Jun-25 at 11:21Yes, the behavior you observe is the expected behavior.
The implementation of HashMap
expects you to use a reasonable hashCode
for the keys. It assumes that your hashCode
would distribute the keys as evenly as possible among the available buckets. If you fail to do that (as you did in your example - where all the keys have the same hashCode
), you will get bad performance.
Under the assumption of even distribution, it makes sense for the HashMap
to double its size once you pass the load factor. It doesn't check how many buckets are actually empty (since it has no way of knowing if new entries will be assigned to empty buckets or to occupied buckets). It just checks the average number of entries per bucket. Once that number exceeds the load factor, the number of buckets is doubled.
QUESTION
I am getting a user input from two text fields and storing them into a hash map. I have checked to see if the values are being inserted correctly to the hash map. and they are. once I try to pass it using Intent.putExtra, I am getting this error.
...ANSWER
Answered 2017-Mar-08 at 09:56use this
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No vulnerabilities reported
Install hashmapTest
You can use hashmapTest like any standard Java library. Please include the the jar files in your classpath. You can also use any IDE and you can run and debug the hashmapTest component as you would do with any other Java program. Best practice is to use a build tool that supports dependency management such as Maven or Gradle. For Maven installation, please refer maven.apache.org. For Gradle installation, please refer gradle.org .
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