jeyaraj

• Sep 29th, 2011

 

(Note: If both the p and P specified in the code are same then)

It prints the Size of the Value of pointer and not the Size of the pointer.

And the value it prints is always '1'

amitabha

• Oct 7th, 2011

 

char *p ....it holds the address of a character.. so a address always is a integer.. hence the answer is 2.

visweswararao

• Oct 27th, 2011

 

here char *p p is a pointer not *p is a pointer
the sizeof(*p) is 1 byte
the sizeof(p) is 2 byte because it is a pointer so it always takes 2 bytes of memory

surendrap

 Profile Answers by surendrap 

• Nov 16th, 2011

 

the sizeof(*p) will be 1 because here p is a pointer variable of type char . and size of char is 1.

i.e. sizeof(*p)= 1.

neha

• Nov 20th, 2011

 

size of p is machine dependent in this case

pavitra

• Dec 3rd, 2011

 

The pointer will be independent upon data type.
so it will be take 2 bytes for any data type

surendrap

 Profile Answers by surendrap 

• Dec 4th, 2011

 

pavitra here p is a pointer not *p. here *p is a char variable.
u will be able to know by printing sizeof(p).

yogita.

 Profile Answers by yogita. 

• Jan 2nd, 2012

 

Any pointer has 4 bytes as its size

krishna1

 Profile Answers by krishna1 

• Jan 2nd, 2012

 

If you declare in pointer any of the datatype..its size 2 byte.

yogita.

 Profile Answers by yogita. 

• Jan 3rd, 2012

 

The size of p is 4 bytes,actually it depends on the compiler on which you run this code,if compiler is 32 bit pointer is 4 bytes if compiler is 16 bit pointer size is 2 byte.The size of *p is 1 byte as it holds the value of a character.

Code
void main()


{


  char *p;


  char x = 1;


  p = &x;


 


  printf("The size of pointer p is %d


", sizeof(p));


  printf("The size of *p is %d


",sizeof(*p));


}


 


 

Divya

• Jan 10th, 2012

 

Array[0]

vaishali

• Feb 28th, 2012

 

Size of *p is 1 byte, as it is character pointer.