Answer: Insertion steganography
Explanation: Insertion steganography is the way of encrypting the data with the help of a regular files and message. It is encrypted by the ordinary files because the identification of files can be neglected. This process is carried out just for the protection purpose in extra form and gets decrypted in the destination port .It has the working based on the replacement of the bits in a file .
Explanation:
a. int foo+; (foo+ is an invalid identifier because + is not a valid char in identifiers)
b. foo int; (Syntax error is any error where the syntax is invalid - either due to misplacement of words, bad spelling, missing semicolons etc.)
c. Static semantic error are logical errors. for e.g passing float as index of an array - arr[1.5] should be a SSE.
d. I think exceptions like NullReferenceException might be an example of DME. Not completely sure but in covariant returns that raise an exception at compile time (in some languages) might also come in this category. Also, passing the wrong type of object in another object (like passing a Cat in a Person object at runtime might qualify for DME.) Simplest example would be trying to access an index that is out of bounds of the array.
Answer:
A) It simplifies the process by providing a single user interface for multiple software tools.
Explanation:
Answer:
Voter registration lists, also called voter rolls, are the gateway to voting because a citizen typically cannot cast a vote that will count unless his or her name appears on the voter registration rolls. State and local officials regularly remove—or purge—citizens from voter rolls. In fact, 39 states and the District of Columbiareported purging more than 13 million voters from registration rolls between 2004 and 2006.
Explanation:
Answer:
B. 1 6 3
Explanation:
Given function definition for calc:
void calc (int a, int& b)
{
int c;
c = a + 2;
a = a * 3;
b = c + a;
}
Function invocation:
x = 1;
y = 2;
z = 3;
calc(x, y);
cout << x << " " << y << " " << z << endl;
- Since x is passed by value, its value remains 1.
- y is passed by reference to the function calc(x,y);
Tracing the function execution:
c=3
a=3
b=c+a = 6;
But b actually corresponds to y. So y=6 after function call.
- Since z is not involved in function call, its value remain 3.
So output: 1 6 3
