In the following code, what will be printed? Why? (Code needed to make this a complete program intentionally left out

Computer Science Sep 03, 2020

In the following code, what will be printed? Why? (Code needed to make this a complete program intentionally left out.)

int j;
while (j < 3)
{
printf('The value of j = %d',j);
j = j + 1;
}
printf('Done!');

Expert Solution

The printout will be:
0
1
2
Done!

Because the integer variable j is not initialized, it will automatically take up the starting value of zero. It will go up to 2 in increments of 1 and print the values out, but as soon as it becomes 3, the condition (3<3) is no longer true and the loop is exited. It then prints Done!.

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