The Clapping Game

A question asked on every pass

You already have both halves of this. A for loop repeats a block. An if picks between two blocks. Put the if inside the loop's braces and the question is no longer asked once — it is asked again for every number the loop counts up to, and the answer can come out differently each time.

Think of an inspector walking down a train carriage. Passing every seat is the loop; glancing at the ticket is the question. Most people are left alone, but the glance happens at every seat without exception.

Braces are what put it inside

Until now the loop's body was a single printf call. Here it is a whole if/else, and that block in turn has two bodies of its own.

for (int count = 1; count <= 20; count++) {
    if (count == 3) {
        printf("clap\n");
    } else {
        printf("%d\n", count);
    }
}
printf("finished\n");

Read it by which braces a line sits inside:

• Inside the for's { } means runs every pass — twenty times here.
• Inside the if's or else's { } means runs only when that branch is chosen, and only on the passes where it is chosen.
• Outside every brace means runs once, after the loop has finished entirely. printf("finished\n"); is not part of the loop at all.

Indentation is there for you to read, not for the compiler to obey — but keeping it honest, as the examples here do, is what makes the braces easy to check by eye.

Every third, not the third one

count == 3 is true on exactly one pass. To catch 3, 6, 9, 12 and the rest you need something they all share: divide any of them by 3 and nothing is left over.

% from the party planner measures that leftover. 7 % 3 is 1, because two threes fit into seven and one is left behind. 9 % 3 is 0, because three threes fit exactly. A leftover of zero is the mark of every third number, so asking whether the leftover equals zero is the test you want.

A worked example

A row of seats, not a clapping game, so the answer to this exercise stays yours to write:

#include <stdio.h>

int main(void) {
    for (int seat = 1; seat <= 4; seat++) {
        if (seat % 2 == 0) {
            printf("%d aisle\n", seat);
        } else {
            printf("%d window\n", seat);
        }
    }
    printf("all seated\n");
    return 0;
}

Running that prints five lines:

1 window
2 aisle
3 window
4 aisle
all seated

Four numbers, four decisions. Seats 2 and 4 leave nothing over when divided by two, so they take the first branch; 1 and 3 take the second. The last line sits outside every brace, so it appears once at the bottom rather than after every seat.

Your turn

Press Run before changing anything. The count from 1 to 20 prints fine, but there is a single clap at 3 and then nothing but numbers, because the test currently reads:

if (count == 3) {

Replace it with a test that is true for every third count: use % and compare the leftover with 0. The loop, the two printf calls and the banners are finished — that one test is the only thing to touch.

When it works the claps land on 3, 6, 9, 12, 15 and 18, and the beat is visible from across the room.

If something goes wrong

If the claps come out on 1, 4, 7 and 10 instead of 3, 6, 9 and 12, the test is comparing the leftover with the wrong number. Leftovers after threes go 1, 2, 0, 1, 2, 0, and it is the 0 that marks every third count.

If the compiler stops on the if line complaining that the left-hand side of an assignment is not assignable, count the equals signs: one stores a value, two ask a question, and only the question belongs inside an if's parentheses.

If the claps vanish entirely and only numbers print, check that the if is still inside the for loop's own braces — one misplaced } is enough to move it outside, where count no longer changes from pass to pass the way you expect.

None of these break anything. The program either runs with an odd rhythm or refuses to compile, and either way the fix is a single line away.