Functions, in one page

Worked example 1 · the lecture's in-class activity

Temperature converter, with no logic in main

"Write celsiusToFahrenheit(double c), returning c * 9.0/5.0 + 32. Write printTemperatureReport(double c), a void function that calls the function above and prints both values nicely. In main(), declare the prototypes, then call printTemperatureReport() for three different temperatures. Trace what happens if you accidentally write void celsiusToFahrenheit(...) with no return."

1. Answer the three questions for each function. celsiusToFahrenheit: in, one double; out, one double; "returning", so return. printTemperatureReport: in, one double; out, nothing; "prints", so void.
2. Write both prototypes at the top, copied from those answers, each ending in ;.
3. Write main as three calls and nothing else. Pick test temperatures you can check by hand: 0 (freezing: 32 °F), 37 (body temperature: 98.6 °F), and −40, the one temperature that is the same on both scales.
4. Write the definitions below main. The report function calls the converter, catches its return value in a variable, and prints. A function calling a function is completely normal.

#include <stdio.h>

double celsiusToFahrenheit(double c);
void printTemperatureReport(double c);

int main(void) {
    printTemperatureReport(0.0);
    printTemperatureReport(37.0);
    printTemperatureReport(-40.0);
    return 0;
}

double celsiusToFahrenheit(double c) {
    return c * 9.0 / 5.0 + 32;
}

void printTemperatureReport(double c) {
    double f = celsiusToFahrenheit(c);
    printf("%7.2f C = %7.2f F\n", c, f);
}

   0.00 C =   32.00 F
  37.00 C =   98.60 F
 -40.00 C =  -40.00 F

5. Check one line by hand. 37 × 9 = 333; 333 ÷ 5 = 66.6; 66.6 + 32 = 98.6 ✓.
6. The activity's trap question. Change the converter's return type to void (in the prototype and the definition) and keep everything else. Clang refuses with two errors, one per misuse:

   error: void function 'celsiusToFahrenheit' should not return a value [-Wreturn-mismatch]
   error: initializing 'double' with an expression of incompatible type 'void'

   The first is inside the converter: a void function can't hand back c * 9.0/5.0 + 32. The second is in the report function: there's nothing coming back to store in f. Both say the same thing: the header's return type must match what the function does and how it's used.

Worked example 2 · predict the output

A global, a parameter with the same name, and a local that shadows it

#include <stdio.h>

int count = 5;                       /* global */

void bump(int count) {               /* parameter hides the global */
    count = count + 10;
    printf("bump: %d\n", count);
}

void addOne(void) {
    count++;                         /* no local count here: the global */
}

int main(void) {
    int x = count;
    bump(x);
    addOne();
    printf("main: %d %d\n", x, count);
    int count = 100;                 /* from here on, main's count hides the global */
    addOne();
    printf("main: %d\n", count);
    return 0;
}

1. For every name, ask "which box?" Inside a function, a name means the nearest declaration: a local or parameter of that function if there is one, otherwise the global. A local only counts from the line where it's declared onwards.
2. Draw one column per box, not per name. There are three different counts here: the global, bump's parameter, and (later) main's local.
3. Walk main line by line, stepping into each call.

   line	global count	main's x	bump's count	main's count	prints
   start	5	—	—	—
   int x = count;	5	5	—	—	(no local count yet → global)
   bump(x)	5	5	5 → 15	—	bump: 15
   addOne()	6	5	gone	—
   printf	6	5	—	—	main: 5 6
   int count = 100;	6	5	—	100
   addOne()	7	5	—	100	(addOne can't see main's locals)
   printf	7	5	—	100	main: 100

4. Read off the output (verified by compiling and running):

   bump: 15
   main: 5 6
   main: 100

Three things to notice. bump changed only its own parameter, so the global stayed 5 during that call. addOne reaches the global both times, even after main declared its own count, because a local is visible only inside the function that declares it. And the global ends at 7, which is never printed: after int count = 100;, main has no way to name the global any more. Compiled with clang -Wall, this program gives zero warnings. That silence is why "prefer parameters and return values" is the rule.

Worked example 3 · write one from a spec

An isPrime function, used from main

"Write a function that checks whether a number is prime. Use it to print all primes from 1 to 30 and how many there are."

1. Three questions. In: one int. Out: yes/no, so bool (from <stdbool.h>, as in Lab 4). "Checks" means return, not print. Name it as a question: bool isPrime(int n).
2. Split the jobs. isPrime knows nothing about 1 to 30 or about printing. It answers one question about one number. main owns the loop, the printing and the count. That split is the point of the exercise.
3. Body of isPrime: Lab 4's prime test, with return doing the flag's job. Numbers below 2 aren't prime, so return false at once. Then try divisors: the first one found means return false, which leaves the function immediately. If the loop finishes without finding one, the number is prime.
4. Write main as an ordinary counting loop that calls isPrime(k) as its if condition. A call that returns bool can sit wherever a condition can.

#include <stdio.h>
#include <stdbool.h>

bool isPrime(int n);

int main(void) {
    int count = 0;
    for (int k = 1; k <= 30; k++) {
        if (isPrime(k)) {
            printf("%d ", k);
            count++;
        }
    }
    printf("\n%d primes up to 30\n", count);
    return 0;
}

bool isPrime(int n) {
    if (n < 2) return false;             /* 0, 1, negatives */
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0) return false;    /* found a divisor: leave now */
    }
    return true;                         /* loop finished: no divisor */
}

2 3 5 7 11 13 17 19 23 29
10 primes up to 30

5. Test the edges, not just the middle. 1 (not prime: the n < 2 line), 2 (prime: the loop runs zero times because 2 × 2 > 2, so it falls through to return true), 4 (first composite), 29 (largest in range). Ten primes up to 30 matches the list you know.

int maxOfThree(int a, int b, int c) {
    int best = a;
    if (b > best) best = b;
    if (c > best) best = c;
    return best;
}