The following program fragment
int i=10;
void main ( )
{
int i=20;
{
int i=30.;
cout << i << :: i ;
}
}
A. | prints 3010 |
B. | prints 3020 |
C. | will result in a run time error |
D. | none of the above |
Option: A Explanation : :: is basically meant to manipulate a global variable, in case a local variable also has the same name. Click on Discuss to view users comments. |
A function abc is defined as
void abc (int x=0, int y=0)
{
cout << x << y;
}
Which of the following function calls is/are illegal? (Assume h, g are declared as integers)
A. | abc () ; |
B. | abc (h) ; |
C. | abc (g, h) ; |
D. | None of the above |
Option: D Explanation : Both the arguments are optional. All the calls are legal. Click on Discuss to view users comments. |
The following c++ code results in
#include "iostream.h"
void main (void)
{
cout << (int i=5) << (int j=6);
}
A. | compilation error |
B. | run time error |
C. | link time error |
D. | none of the above |
Option: A Explanation : Click on Discuss to view users comments. |
The statements
int a = 5;
cout << "FIRST" << (a <<2) << "SECOND";
outputs
A. | FIRST52SECOND |
B. | FIRST20SECOND |
C. | SECOND25FIRST |
D. | an error message |
Option: B Explanation : The symbol << has a context sensitive meaning. The << in (a << 2) means shifting a by 2 bits to the left, which is nothing but multiplying it by 4. So, a < <2 will be 5 x 4 = 20 and hence the output will be FIRST20SECOND. Click on Discuss to view users comments. |
The following program fragment
void main ( )
{
int x=10;
int &p=x;
cout << &p << &x;
}
A. | prints 10 and the address of x |
B. | prints the address of p twice |
C. | prints the address of x twice |
D. | Both (b) & (c) |
Option: D Explanation : int &p=x aliases p to x. This means they refer to the same memory location. So, the address of x will be same as that of p. Click on Discuss to view users comments. |
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