while: the Idea of Repetition

Bob's rocket game needs a countdown: 3, 2, 1, liftoff. He writes four printf lines and it works.

Then the designer asks for a countdown from 100. Bob does not want to write 101 lines, and he should not. One idea, "print the number, then make it smaller", has to run again and again.

A loop is a piece of code that runs again for as long as a condition allows. Computers never get bored of repeating, which is the whole reason they are useful.

while: the same block, again and again

A while loop tests a condition. If the condition is true, it runs the block once, then goes back and tests again. It stops the first time the test is false.

while (condition) {
    statements
}

#include <stdio.h>

int main(void)
{
    int n = 3;

    while (n > 0) {
        printf("%d\n", n);
        n--;
    }
    printf("liftoff\n");
    return 0;
}

3
2
1
liftoff

Change the 3 to 100 for the long countdown. No other line changes.

So a while is an if that goes back to its own test after every pass. The body repeats until the answer turns into no.

Trace a loop with a table before you run it

The same lines run many times, so a loop is hard to read at a glance. Use a trace table (Module 5), one row per pass. Each row holds the variables at the test, the answer, and what the body did.

Count the rows. The body ran 3 times, but the test ran 4 times. The last test, the one that says false, is the one that ends the loop.

The highlighted arrow is the back edge: the jump from the end of the body back to the test. An if has none, and that arrow is the whole difference between a decision and a loop.

So fill in the table on paper before you run a loop. If the last row never arrives, the loop never ends.

Start, condition, step: the three parts of every loop

Every loop that ends has three parts. The start gives the variable its first value. The condition decides whether another pass runs. The step changes the variable, so the condition can become false.

The empty brackets are an error on every command line, the Playground included: error: expected expression before ')' token.

Note the zero-pass case. A while tests before the first pass, so its body may never run at all. Lesson 3 meets a loop that always runs at least once.

So when a loop misbehaves, check the three parts in order. Can the condition become false, and does the step move towards it?

Bob's timer that never ends

Bob writes a kitchen timer. It should count five seconds down and then beep. He rushes the step.

#include <stdio.h>

int main(void)
{
    int seconds = 5;

    while (seconds > 0) {
        seconds - 1;
    }
    printf("beep\n");
    return 0;
}

The line seconds - 1; works out 4 and throws it away, so seconds stays 5. The loop has no real step.

On the Playground nothing appears. After 10 seconds (the limit for a free or signed-out learner), the Playground stops the program. The status badge reads Time limit exceeded, and there is no output, not even the beep.

A judge, the program that grades problems against hidden tests, uses the same name. It usually allows 1 second per test. When it says Time limit exceeded, suspect a step that never moves towards false.

The Playground compiles Bob's timer in silence. A local gcc -Wall on GCC 12 says warning: statement with no effect [-Wunused-value]. The fix is seconds--;.

So a step has to store its result. A calculation that goes nowhere is not a step.

Stop at a sentinel: read until a 0 arrives

Maria's till reads prices, and the cashier types 0 when the basket is empty. A value like that 0, which means "stop" instead of "data", is a sentinel.

#include <stdio.h>

int main(void)
{
    int price = 0;
    int item = 0;

    while (scanf("%d", &price) == 1 && price != 0) {
        item++;
        printf("item %d: %d\n", item, price);
    }
    printf("%d items\n", item);
    return 0;
}

item 1: 120
item 2: 45
item 3: 300
3 items

That output is for the input 120 45 300 0 99. The 0 is the sentinel, and the 99 after it is never read.

The condition does two jobs, in order. First scanf reads a number and must return 1. Only then does price != 0 run, because && skips its right side when the left is false (Module 4).

The sentinel is read, fails the test, and never reaches the body. That is why "3 items" does not count the 0.

So a sentinel ends the loop and is never processed as data. Put the read and the sentinel check in the same condition.

Read until the input ends

Many judge problems have no sentinel. They say "numbers follow, up to the end of the input", and never say how many. That needs the pattern Module 3 promised.

#include <stdio.h>

int main(void)
{
    int mark = 0;
    int passed = 0;

    while (scanf("%d", &mark) == 1) {
        if (mark >= 40) {
            passed++;
        }
    }
    printf("%d passed\n", passed);
    return 0;
}

3 passed

That output is for the input 72 35 90 on one line and 40 12 on the next. scanf skips spaces and line breaks alike, so the layout of the input does not matter.

Alice's program does not know how many marks are coming. It reads while reading works. After the 12, scanf finds nothing and returns EOF, which is -1 (Module 3, lesson 5). -1 is not 1, so the loop ends.

On the Playground, the end of the input is simply the end of what you typed in the input box. A judge's test file ends the same way. In a terminal on your own computer, press Ctrl+D on Linux and macOS, or Ctrl+Z then Enter on Windows.

With an empty input box the body runs zero times, and the program prints 0 passed. That is the right answer for no marks.

So while (scanf("%d", &x) == 1) means "for every number there is". It also stops at the first thing that is not a number.

Keep the best so far, starting from the first value

Zara logs the temperature every morning and wants the coldest. A loop can keep the best answer so far in a variable, and improve it on each pass.

Her input starts with the number of days. The first reading becomes the coldest so far, because at that moment it is the only one.

#include <stdio.h>

int main(void)
{
    int days = 0;
    int t = 0;
    int coldest = 0;
    int day = 1;

    scanf("%d", &days);
    scanf("%d", &t);
    coldest = t;
    while (day < days) {
        scanf("%d", &t);
        if (t < coldest) {
            coldest = t;
        }
        day++;
    }
    printf("coldest: %d\n", coldest);
    return 0;
}

coldest: 12

That output is for the input 4, then 18 12 25 14. Here day counts the readings taken, and there is always at least one day.

Why not start coldest at 0? Then no morning above freezing could beat it, and this input would print 0, a temperature nobody measured.

So a best-so-far variable starts from a real value, the first one read. Then the answer is always one of the inputs.

Braces on every loop body

Module 5 put braces on every if body, and loops follow the same rule. Without braces, a while owns exactly one statement.

#include <stdio.h>

int main(void)
{
    int n = 3;

    while (n > 0)
        printf("%d\n", n);
        n--;
    return 0;
}

The compiler sees one line in the loop, printf, so the program prints 3 again and again. The Playground is silent. A local gcc -Wall says warning: this 'while' clause does not guard... [-Wmisleading-indentation].

So every loop body in this track gets braces, even a body of one line. Braces, not spaces, tell C where a body ends.

#include <stdio.h>

int main(void)
{
    int p = 1;

    while (p < 100) {
        printf("%d\n", p);
        p = p * 2;
    }
    return 0;
}

1
2
4
8
16
32
64

#include <stdio.h>

int main(void)
{
    int goal = 0;
    int weekly = 0;
    int saved = 0;
    int weeks = 0;

    scanf("%d %d", &goal, &weekly);
    while (saved < goal) {
        saved = saved + weekly;
        weeks++;
    }
    printf("%d weeks, %d saved\n", weeks, saved);
    return 0;
}

7 weeks, 1050 saved

#include <stdio.h>

int main(void)
{
    int x = 0;
    int even = 0;
    int odd = 0;

    while (scanf("%d", &x) == 1) {
        if (x % 2 == 0) {
            even++;
        } else {
            odd++;
        }
    }
    printf("even %d\n", even);
    printf("odd %d\n", odd);
    return 0;
}

even 3
odd 2

while (scanf("%d", &x)) {
    printf("%d\n", x);
}

scanf("%d", &price);
while (price != 0) {
    printf("item: %d\n", price);
    scanf("%d", &price);
}

while (n > 0);
{
    printf("%d\n", n);
    n--;
}

int coldest = 0;
while (day < days) {
    scanf("%d", &t);
    if (t < coldest) {
        coldest = t;
    }
    day++;
}

#include <stdio.h>

int main(void)
{
    int n = 3;

    while (n --> 0) {
        printf("%d\n", n);
    }
    printf("after the loop, n is %d\n", n);
    return 0;
}

#include <stdio.h>

int main(void)
{
    int n = 0;
    scanf("%d", &n);

    /* Start, condition, step: one loop, one number per line. */

    return 0;
}

#include <stdio.h>

int main(void)
{
    int price = 0;
    int total = 0;

    while (scanf("%d", &price) == 1 && price != 0) {
        /* Add this price to the total. */
    }

    /* Print the total. */

    return 0;
}

#include <stdio.h>

int main(void)
{
    int x = 0;
    int count = 0;

    while (scanf("%d", &x) == 1) {
        /* One more reading. */
    }

    /* Print the count. An empty log must still print a number. */

    return 0;
}

#include <stdio.h>

int main(void)
{
    int x = 0;

    while (scanf("%d", &x) == 1) {
        /* Keep the best score so far. What is it before any score arrives? */
    }

    /* Print the best score, or empty if nothing was read. */

    return 0;
}

Next, for puts the start, the condition and the step on one line, and Bob tries to print 1 to 10.