break and continue

#include <stdio.h>

int main(void)
{
    printf("for loop, continue skips to the step: ");
    for (int i = 1; i <= 6; i++) {
        if (i % 2 == 0) {
            continue;
        }
        printf("%d ", i);
    }
    printf("\n");

    printf("while loop, written correctly: ");
    int i = 0;
    while (i < 6) {
        i++;                            /* advance FIRST */
        if (i % 2 == 0) {
            continue;
        }
        printf("%d ", i);
    }
    printf("\n");
    return 0;
}

for loop, continue skips to the step: 1 3 5
while loop, written correctly: 1 3 5

The for loop is safe because i++ is in the header and continue runs it. The while loop was made safe by advancing i before the continue can be reached.

Written the other way round it never ends:

int i = 0;
while (i < 6) {
    if (i % 2 == 0) {
        continue;          /* jumps to the condition; i never changes */
    }
    printf("%d ", i);
    i++;
}

i starts at 0, which is even, so continue runs, the condition is tested, i is still 0, and so on for ever. The rule: in a while loop, make the progress happen before any continue can skip it.

break in a switch inside a loop

Chapter 29 said a break in a case does not leave the loop. Here is the proof.

#include <stdio.h>

int main(void)
{
    printf("with break in the case:\n");
    for (int i = 1; i <= 4; i++) {
        switch (i) {
            case 3:
                printf("  i is 3, and break here leaves the SWITCH\n");
                break;
            default:
                printf("  i is %d\n", i);
                break;
        }
    }

    printf("\nto leave the loop, use a flag:\n");
    int done = 0;
    for (int i = 1; i <= 4 && !done; i++) {
        switch (i) {
            case 3:
                printf("  i is 3, setting done\n");
                done = 1;
                break;
            default:
                printf("  i is %d\n", i);
                break;
        }
    }
    printf("\n");
    return 0;
}

with break in the case:
  i is 1
  i is 2
  i is 3, and break here leaves the SWITCH
  i is 4

to leave the loop, use a flag:
  i is 1
  i is 2
  i is 3, setting done

The first loop ran all four passes. The break in case 3 ended the switch for that pass and the loop carried on, which is almost never what a student writing that code intends.

Three ways out of a loop from inside a switch:

1. A flag, tested in the loop's condition, as above.
2. return, if the loop is the last thing the function does.
3. goto, which chapter 31 covers and which is the honest choice for leaving several nested levels at once.

break in nested loops

break leaves one level. In a nested loop it leaves the inner loop only.

#include <stdio.h>

int main(void)
{
    printf("break leaves only the inner loop:\n");
    for (int i = 1; i <= 3; i++) {
        for (int j = 1; j <= 3; j++) {
            if (j == 2) {
                break;
            }
            printf("  i=%d j=%d\n", i, j);
        }
    }

    printf("\nleaving both, with a flag:\n");
    int stop = 0;
    for (int i = 1; i <= 3 && !stop; i++) {
        for (int j = 1; j <= 3; j++) {
            if (i == 2 && j == 2) {
                stop = 1;
                break;
            }
            printf("  i=%d j=%d\n", i, j);
        }
    }
    return 0;
}

break leaves only the inner loop:
  i=1 j=1
  i=2 j=1
  i=3 j=1

leaving both, with a flag:
  i=1 j=1
  i=1 j=2
  i=1 j=3
  i=2 j=1

The first nest printed j=1 three times: the inner break fired on every outer pass and the outer loop kept going.

Where break and continue are not allowed

• continue in a switch that is not inside a loop is an error. There is no pass to continue.
• break outside any loop or switch is an error.
• break in a switch inside a loop applies to the switch. continue in the same place applies to the loop, because a switch has no passes. That asymmetry is worth a mark: the two keywords do not both stop at the switch.

#include <stdio.h>

int main(void)
{
    for (int i = 1; i <= 4; i++) {
        switch (i) {
            case 2:
                continue;               /* skips to the loop's next pass */
            default:
                break;                  /* leaves the switch only */
        }
        printf("reached the end of pass %d\n", i);
    }
    return 0;
}

reached the end of pass 1
reached the end of pass 3
reached the end of pass 4

Pass 2 printed nothing: continue skipped the rest of the loop body, printf included. The other three passes reached the printf, because their break left only the switch.

What this does NOT mean

break does not leave a function. return does.

break does not leave more than one level. Use a flag, a return or a goto.

continue does not restart the loop from the beginning. It moves to the next pass: the step clause in a for, or the condition in a while.

continue in a while does not run anything at the bottom of the body. That is how a while loop with a continue hangs.

break in a case does not leave the enclosing loop. It leaves the switch.

Neither is bad practice. break in a search and continue as a guard at the top of a body are both clearer than the alternatives. What is unclear is several of them scattered through a long body.

Quick revision

• break leaves the innermost enclosing loop or switch.
• continue ends the current pass and goes on to the next.
• In a for, continue runs the step clause. In a while, it goes straight to the condition.
• In a while, advance the counter before any continue can skip it, or the loop never ends.
• break in a case leaves the switch, not the loop. continue in a case does affect the loop.
• break leaves one level of nesting. Use a flag, a return or a goto for more.
• break never leaves a function; return does.
• The search idiom: found_at = -1 before the loop, set and break on a match.

Test yourself

1. What does break do inside a switch that is inside a for loop?

It ends the switch for that pass. The loop continues with its next pass.

2. Why does this loop never end?

int i = 0;
while (i < 5) {
    if (i == 0) continue;
    printf("%d ", i);
    i++;
}

continue jumps to the condition, so i++ at the bottom of the body is never reached while i is 0. i stays 0 for ever. Advance i before the continue.

3. How many lines does this print?

for (int i = 1; i <= 3; i++)
    for (int j = 1; j <= 3; j++) {
        if (j == 2) break;
        printf("%d %d\n", i, j);
    }

Three. The inner break fires on each outer pass after printing j=1, and the outer loop runs three times.

4. How do you leave both loops of a nest at once?

Set a flag tested by the outer loop's condition, return from the function, or goto a label after the outer loop.

5. What is the difference between break and continue in one line?

break ends the loop; continue ends only the current pass.

6. Is continue allowed in a switch?

Only if the switch is inside a loop, and it then applies to the loop, not the switch. Outside a loop it is an error.

What can be asked on this, and how to answer it

"Explain break and continue with examples." Define both in one line each, give a search using break and a skip using continue, and then the two things that earn the extra marks: continue runs the step clause in a for but not the bottom of a while body, and break in a case leaves the switch rather than the loop.

"What is the difference between break and continue?" break ends the innermost loop or switch immediately. continue abandons the rest of the current pass and proceeds to the next. Add that break also works in a switch while continue in a switch refers to an enclosing loop.

"How do you exit from a nested loop?" break leaves only one level. Use a flag in the outer condition, a return if the loop is the last thing in the function, or a goto to a label after the loops. Say which you would choose and why.

"Trace the output" of a loop with break or continue in it. Write the pass number and the variable's value in a small table and mark where control jumped. That table is the answer; the printed lines fall out of it.