This is a continuation of last post.
Before seeing the answer try to predict it & then see the explanation.
All the Best!!
71) struct aaa{
struct aaa *prev;
int i;
struct aaa *next;
};
main()
{
struct aaa abc,def,ghi,jkl;
int x=100;
abc.i=0;abc.prev=&jkl;
abc.next=&def;
def.i=1;def.prev=&abc;def.next=&ghi;
ghi.i=2;ghi.prev=&def;
ghi.next=&jkl;
jkl.i=3;jkl.prev=&ghi;jkl.next=&abc;
x=abc.next->next->prev->next->i;
printf("%d",x);
}
72) struct point
{
int x;
int y;
};
struct point origin,*pp;
main()
{
pp=&origin;
printf("origin is(%d%d)\n",(*pp).x,(*pp).y);
printf("origin is (%d%d)\n",pp->x,pp->y);
}
73) main()
{
int i=_l_abc(10);
printf("%d\n",--i);
}
int _l_abc(int i)
{
return(i++);
}
74) main()
{
char *p;
int *q;
long *r;
p=q=r=0;
p++;
}
75) main()
{
char c=' ',x,convert(z);
getc(c);
if((c>='a') && (c<='z'))
x=convert(c);
printf("%c",x);
}
convert(z)
{
return z-32;
}
========================================================
71.
Answer:
2
Explanation:
above all statements form a double circular linked list;
abc.next->next->prev->next->i
this one points to "ghi" node the value of at particular node is 2.
72.
Answer:
origin is(0,0)
origin is(0,0)
Explanation:
pp is a pointer to structure. we can access the elements of the structure
either with arrow mark or with indirection operator.
Note:
Since structure point is globally declared x & y are initialized as zeroes
73.
Answer:
9
Explanation:
return(i++) it will first return i and then increments. i.e. 10 will be returned.
74.
Answer:
0001...0002...0004
Explanation:
++ operator when applied to pointers increments address according to their
corresponding data-types.
75.
Answer:
Compiler error
Explanation:
declaration of convert and format of getc() are wrong.
========================================================
Thanks for coming till this point!!
VSG
SQL DBA
Before seeing the answer try to predict it & then see the explanation.
All the Best!!
71) struct aaa{
struct aaa *prev;
int i;
struct aaa *next;
};
main()
{
struct aaa abc,def,ghi,jkl;
int x=100;
abc.i=0;abc.prev=&jkl;
abc.next=&def;
def.i=1;def.prev=&abc;def.next=&ghi;
ghi.i=2;ghi.prev=&def;
ghi.next=&jkl;
jkl.i=3;jkl.prev=&ghi;jkl.next=&abc;
x=abc.next->next->prev->next->i;
printf("%d",x);
}
72) struct point
{
int x;
int y;
};
struct point origin,*pp;
main()
{
pp=&origin;
printf("origin is(%d%d)\n",(*pp).x,(*pp).y);
printf("origin is (%d%d)\n",pp->x,pp->y);
}
73) main()
{
int i=_l_abc(10);
printf("%d\n",--i);
}
int _l_abc(int i)
{
return(i++);
}
74) main()
{
char *p;
int *q;
long *r;
p=q=r=0;
p++;
}
75) main()
{
char c=' ',x,convert(z);
getc(c);
if((c>='a') && (c<='z'))
x=convert(c);
printf("%c",x);
}
convert(z)
{
return z-32;
}
========================================================
71.
Answer:
2
Explanation:
above all statements form a double circular linked list;
abc.next->next->prev->next->i
this one points to "ghi" node the value of at particular node is 2.
72.
Answer:
origin is(0,0)
origin is(0,0)
Explanation:
pp is a pointer to structure. we can access the elements of the structure
either with arrow mark or with indirection operator.
Note:
Since structure point is globally declared x & y are initialized as zeroes
73.
Answer:
9
Explanation:
return(i++) it will first return i and then increments. i.e. 10 will be returned.
74.
Answer:
0001...0002...0004
Explanation:
++ operator when applied to pointers increments address according to their
corresponding data-types.
75.
Answer:
Compiler error
Explanation:
declaration of convert and format of getc() are wrong.
========================================================
Thanks for coming till this point!!
VSG
SQL DBA
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