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Vikentia [17]
3 years ago
15

Key issues from a software security perspective are whether the implemented algorithm correctly solves the specified problem, wh

ether the machine instructions executed correctly represent the high level algorithm specification, and whether the manipulation of data values in variables is valid and meaningful. True or false?
Computers and Technology
1 answer:
yKpoI14uk [10]3 years ago
7 0

Answer:

True is the right answer.

Explanation:

Software security can be defined as the security of the software against any malicious, unethical attack or any hacker, etc.

Security makes software to work properly and maintain its authenticity. Security keeps the data safe which was uploaded on software.  

In order to maintain the security of software, a security professional has to check the implemented algorithm which was used to solve the problem, manipulation of data and many others.

Hence the most appropriate answer is true.

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If variable x has value 2 and y has value 2, what is the value of the following Jack expression?
saul85 [17]

Answer:

The value of the following expression is "2".

Explanation:

Here x has value 2 means x=2,and y has value 2 means y=2.

so (x+3)/(4-y)

  (2+3)/(4-2)

5/2 it gives 2 because /(slash) operator gives the quotient part .The slash    operator divide the left hand operand by the right hand operand.On dividing 5/2 it returns integer value i.e 2.

3 0
3 years ago
Each object in your database application is shown in the
lesya [120]
Table
Not so sure about the answer
4 0
3 years ago
Each wireless network has its own name to identify it, known as service set identity. true false
Dominik [7]
True. The acronym for service set identity is SSID, which you've likely saw before. And like you say, it's simply the unique name given to a wireless network so that we are able to identify what network we are connecting to.
8 0
2 years ago
Type (dog, cat, budgie, lizard, horse, etc.) Create a class that keeps track of the attributes above for pet records at the anim
Alenkinab [10]

Answer:

If you did the exercise with two Dog objects, it was a bit boring, right? After all, we have nothing to separate the dogs from each other and no way of knowing, without looking at the source code, which dog produced which bark.

In the previous article, I mentioned that when you create objects, you call a special method called a constructor. The constructor looks like the class name written as a method. For example, for a Dog class, the constructor would be called Dog().

The special thing about constructors is that they are the path to any new object, so they are a great place to call code that initializes an object with default values. Further, the return value from a constructor method is always an object of the class itself, which is why we can assign the return value of the constructor to a variable of the type of class we create.

However, so far, we have not actually created a constructor at all, so how come we can still call that method?

In many languages, C# included, the language gives you a free and empty constructor without you having to do anything. It is implied that you want a constructor; otherwise there would be no way of using the class for anything, so the languages just assume that you have written one.

This invisible and free constructor is called the default constructor, and, in our example, it will look like this:

public Dog(){ }

Notice that this syntax is very similar to the Speak() method we created earlier, except that we do not explicitly return a value nor do we even declare the return type of the method. As I mentioned earlier, a constructor always returns an instance of the class to which it belongs.

In this case, that is the class Dog, and that is why when we write Dog myDog = new Dog(), we can assign the new object to a variable named myDog which is of type Dog.

So let’s add the default constructor to our Dog class. You can either copy the line above or, in Visual Studio, you can use a shortcut: type ctor and hit Tab twice. It should generate the default constructor for you.

The default constructor doesn’t actually give us anything new because it is now explicitly doing what was done implicitly before. However, it is a method, so we can now add content inside the brackets that will execute whenever we call this constructor. And because the constructor runs as the very first thing in an object’s construction, it is a perfect place to add initialization code.

For example, we could set the Name property of our objects to something by adding code such as this:

public Dog()

{

   this.Name = "Snoopy";

}

This example will set the Name property of any new objects to “Snoopy”.

Of course, that’s not very useful because not all dogs are called “Snoopy”, so instead, let us change the method signature of the constructor so that it accepts a parameter.

The parentheses of methods aren’t just there to look pretty; they serve to contain parameters that we can use to pass values to a method. This function applies to all methods, not just constructors, but let’s do it for a constructor first.

Change the default constructor signature to this:

public Dog(string dogName)

This addition allows us to send a string parameter into the constructor, and that when we do, we can refer to that parameter by the name dogName.

Then, add the following line to the method block:

this.Name = dogName;

This line sets this object’s property Name to the parameter we sent into the constructor.

Note that when you change the constructor’s signature, you get a case of the red squigglies in your Program.cs file.When we add our own explicit constructors, C# and .NET will not implicitly create a default constructor for us. In our Program.cs file, we are still creating the Dog objects using the default parameter-less constructor, which now no longer exists.

To fix this problem, we need to add a parameter to our constructor call in Program.cs. We can, for example, update our object construction line as such:

Dog myDog = new Dog(“Snoopy”);

Doing so will remove the red squigglies and allow you to run the code again. If you leave or set your breakpoint after the last code line, you can look at the Locals panel and verify that your object’s Name property has indeed been? Got it?

5 0
2 years ago
1. Actuators apply mechanical force in the form of pressure to overcome
Harman [31]
The answer is d torque because
3 0
3 years ago
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