Answer is
productivity
Sometimes called the office or personal productivity software,
productivity software is dedicated to producing databases, spreadsheets,
charts, graphs, documents, graphs, digital video and worksheets. Reason behind
the name productivity is due to the fact that it increases productivity in
office work.
Answer:
True.
Explanation:
An expression containing the && and operator is only true if both of the operands present in the expression are true otherwise it will give the result as false if either one of them is false or both of them are false.In the question it states that the expression in true either or both of it's operand are true.
Hence the answer is true.
Answer: setJMenubar
Explanation:Menus that are component of the Menu bar that display the options and tools for any function that can be performed in the system have the connection with the window in the field of java through the command of the setJMenubar. It is the command which is given to execute to bring the menu to the particular frame and get it attached according to the JFrame. Therefore, the correct option is setJMenubar.
Answer:
False
Explanation:
The private member of a class is not accessible by using the Dot notation ,however the private member are those which are not accessible inside the class they are accessible outside the class .The public member are accessible inside the class so they are accessible by using the dot operator .
<u>Following are the example is given below in C++ Language </u>
#include<iostream> // header file
using namespace std;
class Rectangle
{
private:
double r; // private member
public:
double area()
{ return 3.14*r*r;
}
};
int main()
{
Rectangle r1;// creating the object
r1.r = 3.5;
double t= r1.area(); // calling
cout<<" Area is:"<<t;
return 0;
}
Output:
compile time error is generated
<u>The correct program to access the private member of class is given below </u>
#include<iostream> // header file
using namespace std;
class Rectangle
{
private:
double r; // private member
public:
double area()
{
r1=r;
double t2=3.14*r2*r2;
return(t2); // return the value
}
};
int main()
{
Rectangle r1;// creating the object
r1.r = 1.5;
double t= r1.area(); // calling
cout<<" Area is:"<<t;
return 0;
}
Therefore the given statement is False
Answer:
I get 0x55 and this the linking address of the main function.
use this function to see changes:
/* bar6.c */
#include <stdio.h>
char main1;
void p2()
{
printf("0x%X\n", main1);
}
Output is probably 0x0
you can use your original bar6.c with updaated foo.c
char main;
int main() // error because main is already declared
{
p2();
//printf("Main address is 0x%x\n",main);
return 0;
}
Will give u an error
again
int main()
{
char ch = main;
p2(); //some value
printf("Main address is 0x%x\n",main); //some 8 digit number not what printed in p2()
printf("Char value is 0x%x\n",ch); //last two digit of previous line output
return 0;
}
So the pain in P2() gets the linking address of the main function and it is different from address of the function main.
Now char main (uninitialized) in another compilation unit fools the compiler by memory-mapping a function pointer on a char directly, without any conversion: that's undefined behavior. Try char main=12; you'll get a multiply defined symbol main...
Explanation:
