INF1070
	    L?sningsforslag oppgaver uke 8 (20.-25.2.2008)

Oppgave 1

#include <stdio.h>
#include <stdlib.h>

static int digval (char c)
{
  /* Hvilken verdi har sifferet `c?? */
  /* Returner -1 om ikke siffer 0-9, a-z eller A-Z. */

  if ('0'<=c && c<='9') return c-'0';
  if ('a'<=c && c<='z') return c-'a'+10;
  if ('A'<=c && c<='Z') return c-'A'+10;
  return -1;
}

int my_atoib (char s[], int base)
{
  int sign = 1, ix = 0, res = 0;

  while (s[ix] && s[ix]!='-' && (digval(s[ix])<0 || digval(s[ix])>=base))
	 ++ix;
  if (s[ix] == '-') {
    sign = -1;  ++ix;
  }
  while (digval(s[ix])>=0 && digval(s[ix])<base)
    res = base*res + digval(s[ix++]);

  return sign*res;
}


char *data[] = { "2de", " -13", " 1FKZ ", "z", "" };

int main (void)
{
  int data_len = sizeof(data)/sizeof(char*), ix;

  for (ix = 0;  ix < data_len;  ++ix) {
    printf("my_atoib(\"%s\",  2) = %d\n", data[ix], my_atoib(data[ix],  2));
    printf("my_atoib(\"%s\", 10) = %d\n", data[ix], my_atoib(data[ix], 10));
    printf("my_atoib(\"%s\", 12) = %d\n", data[ix], my_atoib(data[ix], 12));
    printf("my_atoib(\"%s\", 16) = %d\n", data[ix], my_atoib(data[ix], 16));
    printf("my_atoib(\"%s\", 36) = %d\n", data[ix], my_atoib(data[ix], 36));
  }
  return 0;
}


Oppgave 2

#include <stdio.h>
#include <stdlib.h>
#include <string.h>


static char digit (int v)
{
  if (v <= 9) return '0'+v;
  return 'a'+v-10;
}


static char *itoa_put (char *buf, int n, int base)
{
  if (n >= base) buf = itoa_put(buf, n/base, base);
  *buf = digit(n%base);
  return buf+1;
}

char *my_itoa (int v, int base)
{
  char res[200], *resp = res;

  if (v < 0) {
    *resp = '-';  ++resp;  v = -v;
  }
  resp = itoa_put(resp, v, base);  *resp = 0;
  return strdup(res);
}


int data[] = { 212, -13, 35, 123177, 0 };
int basetab[] = { 2, 10, 12, 16, 36 };

int main (void)
{
  int data_len = sizeof(data)/sizeof(int), ix, 
    base_len = sizeof(basetab)/sizeof(int), ib;

  for (ix = 0;  ix < data_len;  ++ix) {
    for (ib = 0;  ib < base_len;  ++ib) {
      char *s = my_itoa(data[ix],basetab[ib]);
      printf("my_itoab(%d,%d) = \"%s\"\n", data[ix], basetab[ib], s);
      free(s);
    }
}
  return 0;
}


Oppgave 3

.globl mul10
mul10:
	movl	4(%esp),%eax	# eax = v
	addl	%eax,%eax	# eax = 2v
	addl	%eax,%eax	# eax = 4v
	addl	4(%esp),%eax	# eax = 5v
	addl	%eax,%eax   	# eax = 10v
	ret


Oppgave 4

Funksjonen returnerer alltid 0. Den beregner

	   v+2		     i %eax
	   0-v-1-1 = -(v+2)  i %ecx

f?r den summerer de to tallene og f?r 0.

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