The following program fragment
int x[5][5], i, j;
for (i = 0; i < 5, ++i;)
for(j = 0; j < 5 ; j++)
x[ i ] [ j ] = x[ j ][ i ];
A. | transposes the given matrix x |
B. | makes the given matrix x, symmetric |
C. | doesn't alter the matrix x |
D. | none of the above |
Option: B Explanation : Click on Discuss to view users comments. |
Consider the following program segment
i = 6720;
j = 4;
while (( i % j ) == 0)
{
i = i / j ;
j = j + 1;
}
on termination j will have the value
A. | 4 |
B. | 8 |
C. | 9 |
D. | 6720 |
Option: C Explanation : Click on Discuss to view users comments. |
The output of the following program is
main()
{
static int x[ ] = { 1, 2, 3, 4, 5, 6, 7, 8 };
int i;
for (i = 2; i < 6; ++i )
x[ x[ i ] ] = x[ i ];
for( i =0; i < 8; ++i)
printf("%d", x[ i]);
}
A. | 1 2 3 3 5 5 7 8 |
B. | 1 2 3 4 5 6 7 8 |
C. | 8 7 6 5 4 3 2 1 |
D. | 1 2 3 5 4 6 7 8 |
Option: A Explanation : Click on Discuss to view users comments. |
Consider the following program
mani( )
{
int x = 2, y = 5;
if (x < y)
return ( x = x + y);
else printf ("z1");
printf ("z2");
}
Choose the correct statements
A. | the output is z2 |
B. | the output is z1z2 |
C. | this will result in compilation error |
D. | none of above |
Option: D Explanation : Click on Discuss to view users comments. neha.jain@gmail.com said: (8:34pm on Thursday 6th June 2013)
output will be z1z2
Susrutha said: (12:55am on Sunday 22nd April 2018)
It will not print thing because there is return statement in if condition.
|
The for loop
for( i=0; i<10; ++i)
printf("%d", i & 1);
prints
A. | 0101010101 |
B. | 01111111111 |
C. | 0000000000 |
D. | 111111111 |
Option: A Explanation : The binary representation of odd numbers will have a 1 as the least significant digit. So, an odd number ANDed with 1, produces a 1 . Even number end with 0. So. an even number ANDed with 1, produces a 0. This for loop generates even and odd numbers alternatively. So. it prints alternate 0's and 1's. Click on Discuss to view users comments. |