What will print out?

main()
{
char *p1=“name”;
char *p2;
p2=(char*)malloc(20);
memset (p2, 0, 20);
while(*p2++ = *p1++);
printf(“%s\n”,p2);
}

The pointer p2 value is also increasing with p1 .
*p2++ = *p1++ means copy value of *p1 to *p2 , then increment both addresses (p1,p2) by one , so that they can point to next address . So when the loop exits (ie when address p1 reaches next character to “name” ie null) p2 address also points to next location to “name” . When we try to print string with p2 as starting address , it will try to print string from location after “name” … hence it is null string ….

e.g. :
initially p1 = 2000 (address) , p2 = 3000
*p1 has value “n” ..after 4 increments , loop exits … at that time p1 value will be 2004 , p2 =3004 … the actual result is stored in 3000 - n , 3001 - a , 3002 - m , 3003 -e … we r trying to print from 3004 …. where no data is present … that's why its printing null .

Answer: empty string.

 

What will be printed as the result of the operation below:

main()
{
int x=20,y=35;
x=y++ + x++;
y= ++y + ++x;
printf(“%d%d\n”,x,y)
;
}

Answer : 5794

 

What will be printed as the result of the operation below:

main()
{
int x=5;
printf(“%d,%d,%d\n”,x,x<<2,>>2)
;
}

Answer: 5,20,1

 

What will be printed as the result of the operation below:

#define swap(a,b) a=a+b;b=a-b;a=a-b;
void main()
{
int x=5, y=10;
swap (x,y);
printf(“%d %d\n”,x,y)
; swap2(x,y);
printf(“%d %d\n”,x,y)
; }

int swap2(int a, int b)
{
int temp;
temp=a;
b=a;
a=temp;
return 0;
}

as x = 5 = 0×0000,0101; so x << 2 -< 0×0001,0100 = 20; x >7gt; 2 -> 0×0000,0001 = 1. Therefore, the answer is 5, 20 , 1

the correct answer is
10, 5
5, 10


Answer: 10, 5