Hema

• Dec 1st, 2006

 

It will allocate two bytes of memory

winny gupta

 Profile Answers by winny gupta 

• Feb 15th, 2007

 

2 bytes of memory because the pointer variable, whatever data type it maybe pointing to is always an unsigned integer as the address is always a positive integer, hence requiring 2 bytes of memory

avula chiru

 Profile Answers by avula chiru 

• Sep 30th, 2010

 

int, float, or char any pointer variable takes 2bytes of memory..

prititripathi

 Profile Answers by prititripathi 

• Oct 1st, 2010

 

it will take 2 bytes coz pointer store address which r of type int

saurabh1341

 Profile Answers by saurabh1341 

• Oct 22nd, 2010

 

2 bytes.. because p will contain an address, which is 'int'

prititripathi

 Profile Answers by prititripathi 

• Dec 18th, 2010

 

pointer store address so it will require 2 byte

jbode

 Profile Answers by jbode Questions by jbode

• Aug 23rd, 2011

 

The answer will vary by platform. It could be as little as 2 bytes or more than 8. On the systems I work with, it's typically 4 bytes.

bhanu

• Aug 24th, 2011

 

Code
#include<stdio.h>


#include<conio.h>


void main()


{


char *p;


printf(" %d",*p);


}


 

janani

• Aug 25th, 2011

 

thus 32 bits are allocated if we declare a char *p;

r.rubiya

• Aug 26th, 2011

 

char2

Akshaya

• Aug 27th, 2011

 

2 bytes

anusha

• Aug 28th, 2011

 

Two bytes of memory

pooja

• Aug 30th, 2011

 

1 byte

ankitgujrati27

 Profile Answers by ankitgujrati27 

• Sep 1st, 2011

 

& address is of the type integer, it can't be a character or a float.
& Integer takes two bytes to store So answer would be 2bytes.

maggi

• Sep 3rd, 2011

 

2 bytes

suman

• Sep 3rd, 2011

 

*p means there is no certain amount of memory allocated for that it will take what data you give and there is no limitation like up to this size. There is no waste of memory.

navatha

• Sep 7th, 2011

 

2

shafi

• Sep 7th, 2011

 

depends on the data types....but this answer is 1 byte....

shiva

• Sep 10th, 2011

 

1 byte

ronak

• Sep 10th, 2011

 

2 bytes of memory will be allocated, as pointer always point to the unsigned integer as address are unsigned integer.

SHILPA

• Sep 23rd, 2011

 

it will take 2 bytes. because pointer always take 2 bytes

anupama

• Sep 25th, 2011

 

one byte

Avinash Chourasiya

• Nov 12th, 2011

 

2 bytes

jbode

 Profile Answers by jbode Questions by jbode

• Jan 18th, 2012

 

I know Ive answered this already, but I thought Id include a handy macro definition for getting the size of any type (pointer or otherwise) on your platform (see code insert). FWIW, here are the results on my system (Linux 2.4.20, gcc compiler in C99 mode):

Size of char = 1
Size of char * = 4
Size of unsigned char = 1
Size of unsigned char * = 4
Size of short = 2
Size of short * = 4
Size of unsigned short = 2
Size of unsigned short * = 4
Size of long = 4
Size of long * = 4
Size of unsigned long = 4
Size of unsigned long * = 4
Size of long long = 8
Size of long long * = 4
Size of unsigned long long = 8
Size of unsigned long long * = 4
Size of float = 4
Size of float * = 4
Size of double = 8
Size of double * = 4
Size of long double = 12
Size of long double * = 4
Size of struct {char a; char b; long y;} = 8
Size of struct {long y; char a; char b;} = 8
Size of struct {char a; char b; long y;} * = 4
Size of union {char a; char b; long y;} = 4
Size of union {char a; char b; long y;} * = 4

Code
#include <stdio.h>


 


#if defined(__STDC__)


  #if defined(__STDC_VERSION__)


    #if __STDC_VERSION__ >= 199901L


      #define C99 1


    #endif


  #endif


#endif


 


#ifdef C99


  #define fmt "%zu"


  #define cast 


#else


  #define fmt "%lu"


  #define cast (unsigned long)


#endif


 


#define PRINT_SIZE(t) printf("Size of %-30s = " fmt "


", #t, cast sizeof(t))


 


int main(void)


{


  PRINT_SIZE(char);


  PRINT_SIZE(char *);


  PRINT_SIZE(unsigned char);


  PRINT_SIZE(unsigned char *);


  PRINT_SIZE(short);


  PRINT_SIZE(short *);


  PRINT_SIZE(unsigned short);


  PRINT_SIZE(unsigned short *);


  PRINT_SIZE(long);


  PRINT_SIZE(long *);


  PRINT_SIZE(unsigned long);


  PRINT_SIZE(unsigned long *);


#ifdef C99


  PRINT_SIZE(long long);


  PRINT_SIZE(long long *);


  PRINT_SIZE(unsigned long long);


  PRINT_SIZE(unsigned long long *);


#endif


  PRINT_SIZE(float);


  PRINT_SIZE(float *);


  PRINT_SIZE(double);


  PRINT_SIZE(double *);


#ifdef C99


  PRINT_SIZE(long double);


  PRINT_SIZE(long double *);


#endif


  PRINT_SIZE(struct {char a; char b; long y;});


  PRINT_SIZE(struct {long y; char a; char b;});


  PRINT_SIZE(struct {char a; char b; long y;} *);


  PRINT_SIZE(union {char a; char b; long y;});


  PRINT_SIZE(union {char a; char b; long y;} *);


 


  return 0;


}


 

rasmita sahoo

• May 17th, 2012

 

2 byte for tc & 4 byte for gcc compiler

reetu

• Sep 30th, 2012

 

It allocate 2bytes in memory

sourav punoriyar

 Profile Answers by sourav punoriyar 

• Sep 5th, 2013

 

4 bytes in gcc compiler

Mukesh

• Jan 20th, 2016

 

It depends on how much bit Turbo C software you are using. If it is 32 bit then it will take 4 byte. If it is 64 bit then pointer will take 8 bytes of memory. 

Alejandro

• Mar 5th, 2016

 

It Depends on the architecture which is building on.
- In a 16-bit addressing system, 2 bytes.
- In a 32-bits addressing system 4 bytes.
- In a 64-bits addressing system 8 bytes.
Even in old architectures of 16-bits (80x286), depending if it was a far pointer it would size 32-bits (segment + offset).

Raees Ul Islam

• Nov 25th, 2021

 

Pointer is actually is a reference to memory cell. In 32bit operating system, OS use 4 byte for pointing to any memory cell thats why only 4GB memory can be used in 32bit operating system. On the other hand a 64bit OS use 8byte for addressing.
So size of pointer depends upon OS, In 32bit OS pointer will take 4byte and in 64bit OS pointer will take 8 bytes.
No matter pointer is for char, int, bool, Structure it will remain same for all types