On a machine where pointers are 4 bytes long, what happens when the following code is
executed.
main()
{
int x=0,*p=0;
x++; p++;
printf ("%d and %d\n",x,p);
}
a) 1 and 1 is printed
b) 1 and 4 is printed
c) 4 and 4 is printed
d) causes an exception
CHOOSE THE CORRECT OPTION

• Answer is B
  x value is 1 because it is an integer variable having one increment and p is a integer pointer, integer having 4 bytes then one increment is equal to 4 bytes memory. so Answer is B
• 10 years agoHelpfull: Yes(3) No(0)
• b is right
  
• 10 years agoHelpfull: Yes(2) No(0)
• but the %d used in the question wont cause any prob ???
  i mean address are in bytes.. and also, you dont know the base address of this pointer, how can u say that it will give 4, according to me D is right...
• 10 years agoHelpfull: Yes(2) No(0)
• it doesn't cause any exception i think it is (b) because when we increment a pointer it increased by 4 byte in 32 bit plateform. if initially it is 1200 then after p++ it would be 1204.
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• causes an exception. Because p points to NULL, when increasing p it causes an exception.
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• Ans :b
  x++ => 1
  p => gives the address of the 0 ,
  p++ => adress increments ...pointers are 4 bytes long so . p++ gives 4
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• B is Answer because P++ indicate next memory location which is 4.
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• option b
  b is right
  
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• it will print..
  1 and 4
  with a warring message ::: warning: format ‘%d’ expects argument of type ‘int’, but argument 3 has type ‘int *’ [-Wformat]
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• as pointer is of 4 bytes so it will give.......1 and 4 as o/p
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