Arrays, Pointers, and Strings (cont.) - chapter 6

Arguments to main

#include <stdio.h>

int main(int argc, char *argv[])
{
  int   i;

  printf("argc = %d\n", argc);
  for (i = 0; i < argc; ++i)
    printf("argv[%d] = %s\n", i, argv[i]);
  return 0;
}

args.c

Ragged arrays

#include <stdio.h>

int main(void)
{
  char   a[2][15] = {"abc:", "a is for apple"};
  char   *p[2]    = {"abc:", "a is for apple"};

  printf("%c%c%c %s %s\n%c%c%c %s %s\n",
         a[0][0], a[0][1], a[0][2], a[0], a[1],
         p[0][0], p[0][1], p[0][2], p[0], p[1]);
  return 0;
}

ragged_arrays.c

Output

abc abc: a is for apple
abc abc: a is for apple

Functions as arguments

Compute f(x)², x=m,...,n, f(x) is passed as a parameter

#include <stdio.h>
#include <math.h>

double sum_square(double (*f)(double), int m, int n)
{
  int k;
  double sum = 0.0;

  for (k = m; k <= n; ++k)
    sum += f(k) * f(k);
  return sum;
}

double f(double x)
{
  return 1.0 / x;
}

int main(void)
{
  printf("%s%.7f\n%s%.7f\n",
         " First computation: ", sum_square(sin, 2, 13),
         "Second computation: ", sum_square(f, 1, 10000));
  return 0;
}

sum_square.c

gcc -Wall -pedantic sum_square.c -o sum_square -lm

Output

 First computation: 5.7577885
Second computation: 1.6448341

Notice that 1/k² = ²/6

Find a root of f() by the bisection method

#include <stdio.h>

/* epsilon */
#define   EPS   1e-12

double p(double);
double root(double (*)(double), double, double);

int main(void)
{
  double   x;

  x = root(p, 0.0, 3.0);
  printf("%s%.16f\n%s%.16f\n",
         "Approximate root: ", x,
         "  Function value: ", p(x));
  return 0;
}

double root(double f(double), double a, double b)
{
  double   m = (a + b) / 2.0;     /* midpoint */

  if (f(m) == 0.0 || b - a < EPS)
    return m;
  else if (f(a) * f(m) < 0.0)
    return root(f, a, m);
  else
    return root(f, m, b);
}

/* a polynomial p */
double p(double x)
{
  return (x * x * x * x * x  -  7.0 * x  -  3.0);
}

root.c

Output

Approximate root: 1.7196280914844237
  Function value: -0.0000000000012403

Const and pointers

static const int k = 3;

const int a = 7;
int *p = &a;

const int a = 7;
const int *p = &a;

int a = 7;
int * const p = &a;

int a=7;
const int * const p = &a;

Riddle - Crystal Balls

You have b identical crystal balls. There is a high rise with a f floors. One of the floors could be critical. When a crystal ball is dropped from the critical floor or from a higher floor the ball breaks. You need to find the critical floor with the minimum number of experiments in the worst case. Each drop is counted as one experiment regardless of whether the ball breaks or not.

For example, if you have only one ball, you must start your experiments from the first floor. In the worst case you will end up performing an experiment every floor just to find out that there is no critical floor. Given one ball the minimum number of experiments required in the worst case is f.

Last modified: July 24 2004.