Answer:
C. Byte pair encoding is an example of a lossless transformation because an encoded string can be restored to its original version.
Explanation:
Byte pair encoding is a form of encoding in which the most common pairs of consecutive bytes of data are replaced by a single byte which does not occur within the set of data.
For example, if we has a string ZZaaAb, it can be encoded if the pairs of string ZZ are replaced by X and the second pair by Y. So, our data now becomes XYAb.
To get our original data, that is decode it, we just replace the data with the keys X = ZZ and Y = aa thus allowing our original data to be restored.
Since our original string is restored without loss of data, it implies that <u>byte pair encoding is an example of a lossless transformation because an encoded string can be restored to its original version.</u>
Answer: hello your question is poorly written and I have been able to properly arrange them with the correct matching
answer
Static libraries : C
Dynamic link libraries: A
Using static libraries: B
Making some changes to DLL: D
Explanation:
Matching each term with its meaning
<u>Static Libraries </u> : Are attached to the application at the compile time using the Linker ( C )
<u>Dynamic link libraries</u> ( DLL ) : Is Loaded at runtime as applications need them ( A )
<u>Using static Libraries </u>: Makes your program files larger compared to using DLL ( B )
<u>Making some changes to DLL </u>: Does not require application using them to recompile ( D )
Answer:
the answer is true. it is a high level language
Answer:
The answer to this question is DVD and DVD-RW
Explanation:
Answer:
Check the explanation
Explanation:
Here is the program with function definition and two sample calls.
Code:
#include <iostream>
using namespace std;
//checkMe FUNCTION which takes values a, b and c
void checkMe(char &a, int &b, int &c)
{
//if sum of b and c is negative and a is 'n', b and c are set to 0, otherwise a is set to 'p'
if((b+c)<0 && a=='n')
{
b = 0;
c = 0;
}
else
{
a = 'p';
}
}
int main()
{
//first test case when else part is executed
char a = 'n';
int b = 5;
int c = 6;
checkMe(a, b, c);
cout<<a<<" "<<b<<" "<<c<<endl;
//second test case when if part is executed
a = 'n';
b = -4;
c = -5;
checkMe(a, b, c);
cout<<a<<" "<<b<<" "<<c<<endl;
return 0;
}
Kindly check the Output below: