Question

int i = 2;
int j = 3;
int n = 92,
while(i <= n) {
printf(num+" ");
num=(i^j)+1;
}​

Answer 1

Answer:

while (i <n)

{

printf(num +"");

num=(i^j)+1;

}

Answer 2

Answer:

This code is a while loop that starts with initial values of i=2, j=3, and n=92.

Step-by-step explanation:

The loop will continue to execute as long as the value of i is less than or equal to n.

Inside the loop, it prints the value of a variable "num" followed by a space, then updates the value of "num" to the result of (i^j)+1, where "^" is the bitwise XOR operator.

The "^" operator compares each bit of the first operand to the corresponding bit of the second operand. If the bits are the same, the corresponding result bit is 0. If the bits are different, the corresponding result bit is 1.

Finally, the value of i is not incremented in the loop, so the loop will run indefinitely.

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