If we declare a pointer like char *p;
how much memory is allocated for pointer variable 'p'.

It will allocate two bytes of memory

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

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

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

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

pointer store address so it will require 2 byte

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.

Code
#include<stdio.h>
#include<conio.h>
void main()
{
char *p;
printf(" %d",*p);
}
 

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

char2

2 bytes

Two bytes of memory

1 byte

& 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.

2 bytes

*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.

2

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

1 byte

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

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

one byte

2 bytes

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;
}
 

2 byte for tc & 4 byte for gcc compiler

It allocate 2bytes in memory