What will be printed by the following program?
#include <stdio.h>
int main(void) {
int i = 0;
while (i < 5) {
if (i == 3) {
i++;
continue;
}
printf("%d ", i);
i++;
}
return 0;
}When i=3, i is incremented to 4 and 'continue' skips printing 3. For i=0, 1, 2, and 4, it prints the number normally.
Consider the following code snippet:
#include <stdio.h>
int main(void) {
int c = 2;
switch (c) {
case 1: printf("A"); break;
case 2: printf("B");
case 3: printf("C"); break;
default: printf("D");
}
return 0;
}Matching case 2 prints 'B'. Since there is no 'break' in case 2, it falls through to case 3, prints 'C', and then hits 'break'.
What is the output of the following C code?
#include <stdio.h>
int main(void) {
int i = 0;
do {
printf("%d ", i);
i++;
} while (i < 0);
return 0;
}A do-while loop executes its body at least once before evaluating the condition. It prints '0 ' and then terminates because i=1 is not < 0.
What will be the output of the C program below?
#include <stdio.h>
int main(void) {
int k = 1;
if (k == 1)
if (k == 2)
printf("Two");
else
printf("One");
return 0;
}In C, an 'else' binds to the nearest preceding unclosed 'if'. Here, the 'else' belongs to 'if (k == 2)'. Since k=1, inner if condition fails and executes else, printing 'One'.
What is the return value/behavior of the following loop?
#include <stdio.h>
int main(void) {
int a = 10;
while (a --> 0) {
;
}
printf("%d", a);
return 0;
}The '-->' is post-decrement followed by greater-than operator (a-- > 0). The loop runs for a=10 down to 1. On a=0, 0 > 0 is false, 'a' decrements to -1, and loop terminates.