Java And Javascript Difference In English

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This is a low level, compiled, procedural language. Java is an object-oriented, high-level, and interpreted language. Java uses objects, while C uses functions. Java is easier to learn and use because it is high level, while C can do more and run faster because it is closer to machine code.

When the question "which language is better?" "As it turns out, the obvious answer is" it depends. "Naturally, you want to make sure that you invest your time and resources in learning a language that you can be successful in and achieve one of the best high-tech jobs with.

When choosing between Java and C, you have to choose between two of the oldest programming languages ‚Äã‚Äãstill widely used and great public. So the best question to ask is: what are the differences between Java and C and who me should I choose

Let’s discuss why the history of these two languages ‚Äã‚Äãis an an important part of the history of computing in general, the differences between Java and C and the language is best to use under what circumstances.

C: a start

C was developed in 1972 for use with the UNIX operating system, which at the era was developed by Bell Labs. It was a pivotal moment in computer history as every operating system since then has its roots in UNIX (which in turn was partially written in C). this mean s that C is present or has influenced almost all software in use today.

C had an incredible influence even on the languages ‚Äã‚Äãthat followed it, including Java. Although C is still used today for scripting and other behind-the-scenes activities, language derivations such as C++ or C# are now more widely used for applications and programs used by the general public.

Java: The Challenger

Java, it was developed by Sun Microsystems in the years between 1991 and 1995; to the origin must be integrated into digital televisions. It can now run on a dizzying variety of platforms . According to Oracle, Java is installed on three billion devices. The most popular platform for Java development is on Android devices, all of which use code written in the Java language. How Java relates to JavaScript? Fortunately, there is an easy answer, for nothing. When JavaScript was developed by Netscape for web browser scripts (then LiveScript), Java has been more and more popularity. In seeing a trend, Netscape launched "Java" in the name, which causes some confusion for years. Rest assured , you don’t need to know JavaScript to learn Java or vice versa

Java vs C: a comparison

Let’s take a look at the two languages ‚Äã‚Äãand compare them on different fronts.

Java vs C: object-oriented vs procedural

An essential difference between these two languages ‚Äã‚Äãis that Java is an object oriented programming language in C is a language procedural. Does this mean to you as a programmer? It is a question whose paradigm every use of language to model work with a problem.

Java breaks the world down into objects: everything is modeled as a thing - an object - with certain qualities and abilities. The objects are reusable and malleable, which mean s you don’t have to constantly reinvent the wheel as you write. These objects then interact with each other in specific ways based on their identity. A shaped object, for example, will not be a smell, but will have a color.

C, on the other hand, is a procedural language, mean ing that it will put data through different processes (which are called procedures or functions) while the program is running . A procedure program will start at the top and move downward in a linear fashion . Although the code can redirect to different points in the program, it still has to follow a specified sequence. You can almost think of it as Choose Your Adventure book, following the directions based on the data.

No way of working is inherently better than the other. On the contrary, it is a question of what makes sense for the problems you face and (possibly - perhaps more importantly ) what makes for you the most sense when you are trying to solve a problem.

C vs Java: low level vs level high

This is a low level language . This mean s that your interaction with your computer by typing C is the closest to the machine code (ones and zeros) found at the lowest level of the machine. It still uses a syntax that you will recognize with the English language, but it has less abstraction or distance from machine code

Java is a high level language. It has syntax that is more remote or subtracted from machine instructions . it is closer to human language. For this reason, Java is generally easier to learn and use.

If programming languages ‚Äã‚Äãwere compared to human languages, a high level programming language would be like Spanish to English. Without training it may seem difficult to understand except for a few words, but Spanish is still relatively easy to learn as an English speaker. Low-level programming languages are further removed from our natural language and closer to machine code, just as Chinese is much further from English.

Because they work closer to the actual machine code of the computer. Lower level languages ‚Äã‚Äãare generally considered more difficult to read than higher level ones, but more effective because they work more closely with the material.

another way of looking is:. Higher level language is generally more human readable , but at the expense of a lot of machines and activities going on under the surface wordy and difficult to follow at first, but you are aware of every bit.

Java vs C: Interpreted vs Compiled

There is another very important point of difference that we will need in the guts of how computers read a program you write: Java is a semi-interpreted language - using the Java Virtual Machine (JVM) - while C is a compiled language

Let’s find out by examining how we communicate. with a computer. Every time you write a program, it doesn’t matter what language, high or low level - it is always interpreted in a certain way by the processor. In other words, it has multiple translation steps that it must go through to run the program. An interpreted program is like having a native speaker translate a text for you as you read it, while a compiled program is like having the entire text sent to you already translated.

What does this mean to you as a programmer? Languages ‚Äã‚Äãhave compiled a reputation for being faster and more efficient than interpreted languages ‚Äã‚Äãand allow lower level control, such as memory management . As the name suggests, they must be compiled - the program must be compiled by the compiler - before they can be executed. Which is an extra step every time you make a change to your program.

Interpreted languages ‚Äã‚Äãcan be executed after any change without this additional step. Just change what you want and run it. The trade-off between comes with a slightly slower performance; although this becomes less true over time as languages ‚Äã‚Äãare interpreted faster and faster.

C vs Java: memory

One big task that C programmers have to handle on their own is memory management . When manipulating data in a C program, the developer should use calls like "malloc" (memory allocation) and "free" to delimit the memory their program will use. Java uses what is called a garbage collector to purge all the objects that are no longer in use. In other words, Java manages the memory for you. Think of it a bit like driving a manual transmission versus driving an automatic. If you don’t want to worry about shifting gears and prefer to just focus on driving, you will get an automatic. But if you like the control that comes with handling shifting alone (eg, in bad weather), you’d prefer a manual transmission. If this is your first language, Java may be a little more accessible in this regard.

Java vs C speed: which is the most fast

If it’s a matter of speed concern between Java and C, it’s a little hard to compare apples to apples. One thing that keeps C going is that you can write code that is extremely lean with little overhead. This is because it is a lower level language, which mean s that it does not need the same syntactic baggage required by a higher level language. In addition, C has not integrated garbage collector to slow his race and language is compiled, it mean s that the language should not be interpreted in place.

Java is compiled into a lower language and then interpreted. It also has automatic waste collection and is primarily the furthest from machine code. By reason of this code C has a tendency to run faster than Java, but the difference depends on what you do and how the code is optimized.

what should you learn first: Java or C

Learning It is a bit like learning Latin: you will have the basis of a lot of ’ other programming languages, which will make their learning much easier. This is used in many contexts for scripting and other low level tasks.However, there is a good chance that you won’t be writing the next Angry Birds (but with a lot of work, it might ).

Java will be a bit more accessible because it is a higher level language than C. The tradeoff is that you may not fully understand everything that goes on in a program at the expense of brevity and learning speed. However, if you want to dive into what Java does, the mean s are there.

Java programming is much more prevalent in use with programming applications and other programs intended for the public. You’ll likely see more jobs explicitly asking for Java programmers, but those numbers are dropping too. However, Github Java lists as the third most popular language in its repositories , which mean s people are actively using. Although C is listed at number 9 (note that C++ and C# - two very popular derivatives of C - are at 6 and 5, respectively

If you are new to programming you are d naturally want to start with one of the simpler programming languages ‚Äã‚Äã. Neither C nor Java can be described as simple, but they both offer a rich introduction to some important programming concepts.

And the winner is…

If you are looking to learn a computer language to improve your skills, Java and C are excellent candidates and be a great base for continuing to learn other languages.

Learning C will require you to learn more about the ’ under the hood ’ aspects. ¬ªProgramming while you competing in lower level computer and software development tasks. Java is a great introduction to object-oriented programming, which is a widely used programming paradigm.

Part of being a developer is adapting to change, so any of these languages ‚Äã‚Äãwould give you a solid background to learn and adapt to any programming challenges that arise.


Java And Javascript Difference In English __del__: Questions

__del__

How can I make a time delay in Python?

5 answers

I would like to know how to put a time delay in a Python script.

2973

Answer #1

import time
time.sleep(5)   # Delays for 5 seconds. You can also use a float value.

Here is another example where something is run approximately once a minute:

import time
while True:
    print("This prints once a minute.")
    time.sleep(60) # Delay for 1 minute (60 seconds).

2973

Answer #2

You can use the sleep() function in the time module. It can take a float argument for sub-second resolution.

from time import sleep
sleep(0.1) # Time in seconds

__del__

How to delete a file or folder in Python?

5 answers

How do I delete a file or folder in Python?

2639

Answer #1


Path objects from the Python 3.4+ pathlib module also expose these instance methods:

Java And Javascript Difference In English exp: Questions

exp

How do I merge two dictionaries in a single expression (taking union of dictionaries)?

5 answers

Carl Meyer By Carl Meyer

I have two Python dictionaries, and I want to write a single expression that returns these two dictionaries, merged (i.e. taking the union). The update() method would be what I need, if it returned its result instead of modifying a dictionary in-place.

>>> x = {"a": 1, "b": 2}
>>> y = {"b": 10, "c": 11}
>>> z = x.update(y)
>>> print(z)
None
>>> x
{"a": 1, "b": 10, "c": 11}

How can I get that final merged dictionary in z, not x?

(To be extra-clear, the last-one-wins conflict-handling of dict.update() is what I"m looking for as well.)

5839

Answer #1

How can I merge two Python dictionaries in a single expression?

For dictionaries x and y, z becomes a shallowly-merged dictionary with values from y replacing those from x.

  • In Python 3.9.0 or greater (released 17 October 2020): PEP-584, discussed here, was implemented and provides the simplest method:

    z = x | y          # NOTE: 3.9+ ONLY
    
  • In Python 3.5 or greater:

    z = {**x, **y}
    
  • In Python 2, (or 3.4 or lower) write a function:

    def merge_two_dicts(x, y):
        z = x.copy()   # start with keys and values of x
        z.update(y)    # modifies z with keys and values of y
        return z
    

    and now:

    z = merge_two_dicts(x, y)
    

Explanation

Say you have two dictionaries and you want to merge them into a new dictionary without altering the original dictionaries:

x = {"a": 1, "b": 2}
y = {"b": 3, "c": 4}

The desired result is to get a new dictionary (z) with the values merged, and the second dictionary"s values overwriting those from the first.

>>> z
{"a": 1, "b": 3, "c": 4}

A new syntax for this, proposed in PEP 448 and available as of Python 3.5, is

z = {**x, **y}

And it is indeed a single expression.

Note that we can merge in with literal notation as well:

z = {**x, "foo": 1, "bar": 2, **y}

and now:

>>> z
{"a": 1, "b": 3, "foo": 1, "bar": 2, "c": 4}

It is now showing as implemented in the release schedule for 3.5, PEP 478, and it has now made its way into the What"s New in Python 3.5 document.

However, since many organizations are still on Python 2, you may wish to do this in a backward-compatible way. The classically Pythonic way, available in Python 2 and Python 3.0-3.4, is to do this as a two-step process:

z = x.copy()
z.update(y) # which returns None since it mutates z

In both approaches, y will come second and its values will replace x"s values, thus b will point to 3 in our final result.

Not yet on Python 3.5, but want a single expression

If you are not yet on Python 3.5 or need to write backward-compatible code, and you want this in a single expression, the most performant while the correct approach is to put it in a function:

def merge_two_dicts(x, y):
    """Given two dictionaries, merge them into a new dict as a shallow copy."""
    z = x.copy()
    z.update(y)
    return z

and then you have a single expression:

z = merge_two_dicts(x, y)

You can also make a function to merge an arbitrary number of dictionaries, from zero to a very large number:

def merge_dicts(*dict_args):
    """
    Given any number of dictionaries, shallow copy and merge into a new dict,
    precedence goes to key-value pairs in latter dictionaries.
    """
    result = {}
    for dictionary in dict_args:
        result.update(dictionary)
    return result

This function will work in Python 2 and 3 for all dictionaries. e.g. given dictionaries a to g:

z = merge_dicts(a, b, c, d, e, f, g) 

and key-value pairs in g will take precedence over dictionaries a to f, and so on.

Critiques of Other Answers

Don"t use what you see in the formerly accepted answer:

z = dict(x.items() + y.items())

In Python 2, you create two lists in memory for each dict, create a third list in memory with length equal to the length of the first two put together, and then discard all three lists to create the dict. In Python 3, this will fail because you"re adding two dict_items objects together, not two lists -

>>> c = dict(a.items() + b.items())
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for +: "dict_items" and "dict_items"

and you would have to explicitly create them as lists, e.g. z = dict(list(x.items()) + list(y.items())). This is a waste of resources and computation power.

Similarly, taking the union of items() in Python 3 (viewitems() in Python 2.7) will also fail when values are unhashable objects (like lists, for example). Even if your values are hashable, since sets are semantically unordered, the behavior is undefined in regards to precedence. So don"t do this:

>>> c = dict(a.items() | b.items())

This example demonstrates what happens when values are unhashable:

>>> x = {"a": []}
>>> y = {"b": []}
>>> dict(x.items() | y.items())
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unhashable type: "list"

Here"s an example where y should have precedence, but instead the value from x is retained due to the arbitrary order of sets:

>>> x = {"a": 2}
>>> y = {"a": 1}
>>> dict(x.items() | y.items())
{"a": 2}

Another hack you should not use:

z = dict(x, **y)

This uses the dict constructor and is very fast and memory-efficient (even slightly more so than our two-step process) but unless you know precisely what is happening here (that is, the second dict is being passed as keyword arguments to the dict constructor), it"s difficult to read, it"s not the intended usage, and so it is not Pythonic.

Here"s an example of the usage being remediated in django.

Dictionaries are intended to take hashable keys (e.g. frozensets or tuples), but this method fails in Python 3 when keys are not strings.

>>> c = dict(a, **b)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: keyword arguments must be strings

From the mailing list, Guido van Rossum, the creator of the language, wrote:

I am fine with declaring dict({}, **{1:3}) illegal, since after all it is abuse of the ** mechanism.

and

Apparently dict(x, **y) is going around as "cool hack" for "call x.update(y) and return x". Personally, I find it more despicable than cool.

It is my understanding (as well as the understanding of the creator of the language) that the intended usage for dict(**y) is for creating dictionaries for readability purposes, e.g.:

dict(a=1, b=10, c=11)

instead of

{"a": 1, "b": 10, "c": 11}

Response to comments

Despite what Guido says, dict(x, **y) is in line with the dict specification, which btw. works for both Python 2 and 3. The fact that this only works for string keys is a direct consequence of how keyword parameters work and not a short-coming of dict. Nor is using the ** operator in this place an abuse of the mechanism, in fact, ** was designed precisely to pass dictionaries as keywords.

Again, it doesn"t work for 3 when keys are not strings. The implicit calling contract is that namespaces take ordinary dictionaries, while users must only pass keyword arguments that are strings. All other callables enforced it. dict broke this consistency in Python 2:

>>> foo(**{("a", "b"): None})
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: foo() keywords must be strings
>>> dict(**{("a", "b"): None})
{("a", "b"): None}

This inconsistency was bad given other implementations of Python (PyPy, Jython, IronPython). Thus it was fixed in Python 3, as this usage could be a breaking change.

I submit to you that it is malicious incompetence to intentionally write code that only works in one version of a language or that only works given certain arbitrary constraints.

More comments:

dict(x.items() + y.items()) is still the most readable solution for Python 2. Readability counts.

My response: merge_two_dicts(x, y) actually seems much clearer to me, if we"re actually concerned about readability. And it is not forward compatible, as Python 2 is increasingly deprecated.

{**x, **y} does not seem to handle nested dictionaries. the contents of nested keys are simply overwritten, not merged [...] I ended up being burnt by these answers that do not merge recursively and I was surprised no one mentioned it. In my interpretation of the word "merging" these answers describe "updating one dict with another", and not merging.

Yes. I must refer you back to the question, which is asking for a shallow merge of two dictionaries, with the first"s values being overwritten by the second"s - in a single expression.

Assuming two dictionaries of dictionaries, one might recursively merge them in a single function, but you should be careful not to modify the dictionaries from either source, and the surest way to avoid that is to make a copy when assigning values. As keys must be hashable and are usually therefore immutable, it is pointless to copy them:

from copy import deepcopy

def dict_of_dicts_merge(x, y):
    z = {}
    overlapping_keys = x.keys() & y.keys()
    for key in overlapping_keys:
        z[key] = dict_of_dicts_merge(x[key], y[key])
    for key in x.keys() - overlapping_keys:
        z[key] = deepcopy(x[key])
    for key in y.keys() - overlapping_keys:
        z[key] = deepcopy(y[key])
    return z

Usage:

>>> x = {"a":{1:{}}, "b": {2:{}}}
>>> y = {"b":{10:{}}, "c": {11:{}}}
>>> dict_of_dicts_merge(x, y)
{"b": {2: {}, 10: {}}, "a": {1: {}}, "c": {11: {}}}

Coming up with contingencies for other value types is far beyond the scope of this question, so I will point you at my answer to the canonical question on a "Dictionaries of dictionaries merge".

Less Performant But Correct Ad-hocs

These approaches are less performant, but they will provide correct behavior. They will be much less performant than copy and update or the new unpacking because they iterate through each key-value pair at a higher level of abstraction, but they do respect the order of precedence (latter dictionaries have precedence)

You can also chain the dictionaries manually inside a dict comprehension:

{k: v for d in dicts for k, v in d.items()} # iteritems in Python 2.7

or in Python 2.6 (and perhaps as early as 2.4 when generator expressions were introduced):

dict((k, v) for d in dicts for k, v in d.items()) # iteritems in Python 2

itertools.chain will chain the iterators over the key-value pairs in the correct order:

from itertools import chain
z = dict(chain(x.items(), y.items())) # iteritems in Python 2

Performance Analysis

I"m only going to do the performance analysis of the usages known to behave correctly. (Self-contained so you can copy and paste yourself.)

from timeit import repeat
from itertools import chain

x = dict.fromkeys("abcdefg")
y = dict.fromkeys("efghijk")

def merge_two_dicts(x, y):
    z = x.copy()
    z.update(y)
    return z

min(repeat(lambda: {**x, **y}))
min(repeat(lambda: merge_two_dicts(x, y)))
min(repeat(lambda: {k: v for d in (x, y) for k, v in d.items()}))
min(repeat(lambda: dict(chain(x.items(), y.items()))))
min(repeat(lambda: dict(item for d in (x, y) for item in d.items())))

In Python 3.8.1, NixOS:

>>> min(repeat(lambda: {**x, **y}))
1.0804965235292912
>>> min(repeat(lambda: merge_two_dicts(x, y)))
1.636518670246005
>>> min(repeat(lambda: {k: v for d in (x, y) for k, v in d.items()}))
3.1779992282390594
>>> min(repeat(lambda: dict(chain(x.items(), y.items()))))
2.740647904574871
>>> min(repeat(lambda: dict(item for d in (x, y) for item in d.items())))
4.266070580109954
$ uname -a
Linux nixos 4.19.113 #1-NixOS SMP Wed Mar 25 07:06:15 UTC 2020 x86_64 GNU/Linux

Resources on Dictionaries

5839

Answer #2

In your case, what you can do is:

z = dict(list(x.items()) + list(y.items()))

This will, as you want it, put the final dict in z, and make the value for key b be properly overridden by the second (y) dict"s value:

>>> x = {"a":1, "b": 2}
>>> y = {"b":10, "c": 11}
>>> z = dict(list(x.items()) + list(y.items()))
>>> z
{"a": 1, "c": 11, "b": 10}

If you use Python 2, you can even remove the list() calls. To create z:

>>> z = dict(x.items() + y.items())
>>> z
{"a": 1, "c": 11, "b": 10}

If you use Python version 3.9.0a4 or greater, then you can directly use:

x = {"a":1, "b": 2}
y = {"b":10, "c": 11}
z = x | y
print(z)
{"a": 1, "c": 11, "b": 10}

5839

Answer #3

An alternative:

z = x.copy()
z.update(y)

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