C control structures: if, switch, while, for, do-while

In week 10 you built loops and ifs by hand with BR instructions. In C the same structures are written with keywords like if, while, for; the compiler translates them into the BR patterns you wrote. This page explains the rule of each structure and the points where the exam sets the most traps. The question type is again “What is printed?”.

if and if-else

int score = 72;
if (score >= 85) printf("AA\n");
else if (score >= 70) printf("BB\n");
else if (score >= 50) printf("CC\n");
else printf("FF\n");

BB

The conditions are tested from top to bottom, the block of the first true one runs, and the rest are skipped.

PITFALL · = versus ==. if (x = 5) assigns 5 to x and the value of the expression is 5, always true. The compiler usually gives only a warning. When you see this code in the exam, write the output according to the assignment.

Dangling else

if (a == 0)
    if (b == 0) printf("A\n");
else printf("B\n");

The indentation makes you think the else belongs to the outer if. The rule is not that: the else attaches to the nearest unmatched if. That is, it is the else of the inner if (b == 0). When a = 1 nothing is printed. To make your intent clear, always use { }.

switch

switch (expression) {
    case CONST1: ... break;
    case CONST2: ... break;
    default: ...
}

Execution jumps to whichever case the value of the expression matches, and continues until a break (or until the switch ends). If there is no break, the code of the cases below also runs: fall-through.

int d = 1;
switch (d) {
    case 0: printf("zero ");
    case 1: printf("one ");
    case 2: printf("two ");
}

one two

Case labels must be constants (int or char). You cannot write case x > 5:.

while

int n = 3;
while (n > 0) {
    printf("%d ", n);
    n--;
}
printf("go\n");

3 2 1 go

The condition is tested first. If it is false at the start, the body never runs.

The LC-3 counterpart (n in R0):

LOOP    ADD  R0, R0, #0    ; set the CC according to n
        BRnz DONE          ; exit if n <= 0
        ...                ; body
        ADD  R0, R0, #-1   ; n--
        BRnzp LOOP
DONE    ...

for

for (init; condition; update) body;

It is the same as:

init;
while (condition) { body; update; }

int i;
for (i = 1; i <= 4; i++)
    printf("%d ", i * i);

1 4 9 16

The trace table

Solve loop questions with a table:

for (i = 10; i > 0; i /= 2) c++;

i     i > 0?   c
10    yes      1
5     yes      2
2     yes      3
1     yes      4
0     no       (ends)

PITFALL · a semicolon after the for. for (i = 0; i < 5; i++); is a loop with an empty body. The line below, whatever its indentation, runs once after the loop ends and sees i = 5.

do-while

do {
    body;
} while (condition);

The body runs first, the condition is tested afterwards. So it runs at least once. It is ideal for menu loops (“ask until the user enters 0”). The ; at the end is required.

break and continue

• break: leave the innermost loop or switch immediately.
• continue: skip the rest of this round and move on to the next one. In a for, the update part (i++) still runs.

for (i = 0; i < 10; i++) {
    if (i % 3 == 0) continue;   /* 0, 3, 6 are skipped */
    if (i > 7) break;           /* leave at 8 */
    printf("%d ", i);
}

1 2 4 5 7

Nested loops

In each round of the outer loop, the inner loop runs from start to end.

int i, j;
for (i = 1; i <= 3; i++) {
    for (j = 0; j < i; j++)
        printf("*");
    printf("\n");
}

*
**
***

The total number of inner rounds: 1 + 2 + 3 = 6. In “how many times does it run?” questions, write down how many times the inner loop runs for each outer round and add them up.

Worked trace: the Collatz sequence

If n is even halve it, if odd make it 3n + 1; continue until you reach 1.

int n = 6, steps = 0;
while (n != 1) {
    if (n % 2 == 0) n = n / 2;
    else n = 3 * n + 1;
    printf("%d ", n);
    steps++;
}
printf("\n%d\n", steps);

3 10 5 16 8 4 2 1
8

When building the trace table, write three things on each row: the result of the condition, which branch ran, and the new values of the variables.

n    n != 1?   branch     new n   steps
6    yes       even       3       1
3    yes       odd        10      2
10   yes       even       5       3
5    yes       odd        16      4
16   yes       even       8       5
8    yes       even       4       6
4    yes       even       2       7
2    yes       even       1       8
1    no        (ends)

The LC-3 counterpart of the for loop

int s = 0, i;
for (i = 1; i <= 4; i++) s += i;     /* s = 10 */

The compiler (s in R0, i in R1) produces roughly this:

        AND  R0, R0, #0     ; s = 0
        AND  R1, R1, #0
        ADD  R1, R1, #1     ; i = 1          (init)
LOOP    ADD  R2, R1, #-5    ; i - 5
        BRzp DONE           ; exit if i >= 5  (the opposite of the condition)
        ADD  R0, R0, R1     ; s += i          (body)
        ADD  R1, R1, #1     ; i++             (update)
        BRnzp LOOP
DONE    HALT

In the simulator the result is R0 = 10, R1 = 5. The test i <= 4 turns into the question “is i - 5 negative?”: in the LC-3 a comparison always means a subtraction + the condition codes. If the condition is true it falls into the body; if false (z or p) it jumps out of the loop.

Ready-made loop patterns

Common mistakes

int x = 0;
if (x = 5) printf("yes %d\n", x);
else printf("no %d\n", x);

int a = 1, b = 0;
if (a == 0)
    if (b == 0) printf("A\n");
else printf("B\n");
printf("end\n");

int i = 1, s = 0;
while (i <= 10) {
    s += i;
    i += 3;
}
printf("%d %d\n", s, i);

int i;
for (i = 2; i < 15; i += 4)
    printf("%d ", i);

int n = 10;
do {
    printf("%d ", n);
    n++;
} while (n < 5);

int k = 2;
switch (k) {
    case 1: printf("one ");
    case 2: printf("two ");
    case 3: printf("three ");
            break;
    default: printf("other ");
}

int i;
for (i = 0; i < 10; i++) {
    if (i % 3 == 0) continue;
    if (i > 7) break;
    printf("%d ", i);
}

int i, j, c = 0;
for (i = 0; i < 4; i++)
    for (j = i; j < 4; j++)
        c++;
printf("%d\n", c);

int i;
for (i = 0; i < 5; i++);
    printf("%d\n", i);

int i, c = 0;
for (i = 10; i > 0; i /= 2)
    c++;
printf("%d\n", c);

int n = 4096, s = 0;
while (n > 0) {
    s += n % 10;
    n /= 10;
}
printf("%d\n", s);