Code:
#include <stdio.h>
int P1(int *a) {return *a+1;}
int P2(int *b) {return 1+(*b+=1);}
int P3(int *c) {return *c++;}
void main () {
int (*p[]) (int *)={P1,P2,P3,P1,P3,P2};
int i=0,t;
while (1) {
if ((t=(*p[i]) (&i))==i)
break;
printf ("%d",i=t);
}
}
#include <stdio.h>
int P1(int *a) {return *a+1;}
int P2(int *b) {return 1+(*b+=1);}
int P3(int *c) {return *c++;}
void main () {
int (*p[]) (int *)={P1,P2,P3,P1,P3,P2};
int i=0,t;
while (1) {
if ((t=(*p[i]) (&i))==i)
break;
printf ("%d",i=t);
}
}
Code:
#include <stdio.h>
#include <stdlib.h>
void main () {
int i, *a[5], **b;
for (i=0; i<5 ; i++) {
a[i]= (int *) malloc (sizeof(int));
*a[i]=i%2;
}
for (b=a+--i; i>0; i--) {**(b-i)=**b+i%3;}
for(i=0; i<5; i++) printf ("%d", *b[-i]), free (b[-i]);
}
#include <stdio.h>
#include <stdlib.h>
void main () {
int i, *a[5], **b;
for (i=0; i<5 ; i++) {
a[i]= (int *) malloc (sizeof(int));
*a[i]=i%2;
}
for (b=a+--i; i>0; i--) {**(b-i)=**b+i%3;}
for(i=0; i<5; i++) printf ("%d", *b[-i]), free (b[-i]);
}
Resenje prvog zad. je 134, a drugog 01201, medjutim nikako ne mogu da dodjem do ovih rezultata. Nadam se da neko moze da mi pomogne i pojasni ove kodove.