Question

1) Give the output for the following program?

#include <stdio.h>

int i;

int main()
{
for (i=1;i<= 15; i++)
{ printf("%d\n", i);
if (i=10)
break;
}

return 0;

}​

Answer 1

is initialized to 5 and as the loop condition is false before the first iteration itself so the loop doesn't execute. The statement System.out.println(i * 4); is outside the loop so it gets executed once, printing 20 to the console.

★Hope it's helps you...

Answer 2

Explanation:

Step 1: for(;;) this statement will genereate infinite loop.

Step 2: printf("%d\n", i++); this statement will print the value of variable i and increement i by 1(one).

Step 3: if(i>10) here, if the variable i value is greater than 10, then the for loop breaks.

There can exists a switch statement, which has no case.

switch(i) becomes switch(1), then the case 1: block is get executed. Hence it prints "Case1".

printf("This is c program."); is ignored by the compiler.

Hence there is no error and prints "Case1".

The while() loop must have conditional expression or it shows "Expression syntax" error.

Example: while(i > 10){ ... }

Because, case 3 + 2: and case 5: have the same constant value 5.