The following Program
main()
{
int i = 5;
if (i == 5) return;
else printf (" is not five");
printf("over");
}
results in
A. | a syntax error |
B. | an execution error |
C. | printing of over |
D. | execution termination, without printing anything |
Option: D Explanation : Click on Discuss to view users comments. |
The following program fragment
int i = 5;
do
{
putchar(i + 100);
printf("%d", i--);
}
while (i);
results in the printing of
A. | i5h4g3f2el |
B. | i4h3g2fle0 |
C. | an error message |
D. | none of above |
Option: A Explanation : putchar (1 0 5) will print the ASCII equivalent of 105 i.e.. ' i '. The printf statement prints the current value of i. i.e. 5 and then decrements it. So, h4 will be printed in the next pass. This continues until ' i ' becomes 0, at which point the loop gets terminated. Click on Discuss to view users comments. |
The following loop
while(printf("%d", printf("az")))
printf("by");
A. | prints azbybvbvbv... |
B. | an execution error |
C. | prints azbyazbyazbyazby... |
D. | none of the above |
Option: D Explanation : printf("az") prints az and returns a value 2 (since it printed two characters). So. the condition results in the printing of az2. Since it always returns 2, it is an infinite loop. The output will be az2byaz2by. . . Click on Discuss to view users comments. |
The following statements
for (i= 3; i < 15; i += 3)
{
printf ("%d ", i);
++i;
}
will result in the printing of
A. | 3 6 9 12 |
B. | 3 6 9 12 15 |
C. | 3711 |
D. | 3 7 I I 15 |
Option: C Explanation : Click on Discuss to view users comments. |