Skip to main content

Constant Pointer and Pointer to constants


Constant Pointer:-
It is type of pointer whose address cannot be modified after first time initialization.

Ex:-
 #include<stdio.h>
   int main()
   {
       int var = 10, var2 = 20;
       int *const ptr = &var;
       ptr = &var2;                /* error cant change read only ptr is constant*/
       *ptr = 30;                  /* value can be changed */
       printf(“ptr  = %d\n”,*ptr);
      return 0;
  }

In this program ptr is constant which cannot hold the other addresses but value can be modified.

Pointer to a constant:-

It is type of pointer in which the data  pointed cannot be modified.

Ex:-
#include<stdio.h>
     int main()
   {
       int var = 10, var2 = 20;
       const int* ptr = &var;
       ptr = &var2;                /* can change the address */
       *ptr = 30;                  /* error data is constant  */
       printf(“ptr  = %d\n”,*ptr);
      return 0;
  }

In this program ptr points to constant data which cannot be modified,but address can be modified.

Constant pointer to a constant:-

This is combination of above to here data and address are constants.

Ex:-

#include<stdio.h>
   int main()
 {
     int var = 10, var2 = 20;
     const int* const ptr = &var;
      ptr = &var2;                /* error address is constant */
      *ptr = 30;                  /* error data is constant  */
      printf(“ptr  = %d\n”,*ptr);
      return 0;
 }

Easy way to remember:-

const int * ptr :- if * comes after const then data is constant.
int * const  ptr :-  if * comes before const then address is constant.
cons int * const ptr :- address and data both are constant.

Comments

Popular posts from this blog

Sampling and FFT Size derivation in LTE

Sampling and FFT Size derivation in LTE Ts = 1 / (15000 x 2048) seconds, which corresponds to the 30.72 MHz sample clock for the 2048 point FFT used with the 20 MHz system bandwidth. In the frequency domain, the number of sub-carriers N ranges from 128 to 2048, depending on channel bandwidth with 512 and 1024 for 5 and 10 MHz, respectively, being most commonly used in practice. The sub-carrier spacing is ∆f = 1/T u = 15 kHz. The sampling rate is fs = ∆f · N = 15000 N. This results in a sampling rate that’s multiple or sub-multiple of the WCDMA chip rate of 3.84 Mcps: LTE parameters have been chosen such that FFT lengths and sampling rates are easily obtained for all operation modes while at the same time ensuring the easy implementation of dual-mode devices with a common clock reference. Sampling frequency is Multiple's of 2, For 15 Mhz Bandwidth - Sampling Frequency = 23.04 (6 * 3.84). FFT SIZE = S...

C Programming Questions – Part 1

1. W hat do curly braces denote in C? Why does it make sense to use curly brac es to surround the body of a function?   Answer: The curly braces denote a block of code, in which variables can be declared. Variables declared within the block are valid only until the end of the block, marked by the matching right curly brace ’}’. The body of a function is one such type of block, and thus, curly braces are used to describe the extent of that block . 2.Describe the difference between the literal values 7, "7", and ’7 ’ ?   Answer: The first literal is integer 7.Second literal is null terminated string value '7'.Third literal is character '7' having ASCII character code (55). 3. Consider the statement double ans = 10.0+2.0/3.0−2.0∗2.0; Rewrite this statement, inserting parentheses to ensure that ans = 11.0 upon evaluation of this statement ? Answer: double ans = 10.0+2.0/ (( 3.0−2.0 ) ∗2.0 ) ; 4 .C...

Python program to find Matrix Diagonal Difference

Program Statement – Matrix Diagonal Difference You are given a square matrix of size  N × N . Can you calculate the absolute difference of the sums across the main diagonal and the secondary diagonal? https://www.hackerrank.com/challenges/diagonal-difference Program – Python 3 Version mat = [] nrow = int(input()) for i in range(nrow):   mat.extend(list(map(int,input().split()))) #keep extending the array #Main Diagonal Sum indx,pd_sum = 0,0 for i in range(nrow):   pd_sum += mat[indx]   indx += nrow + 1 #Secondary Diagonal Sum indx,sd_sum = nrow-1,0 for i in range(nrow):   sd_sum += mat[indx]   indx += nrow - 1 print(abs(pd_sum – sd_sum)) Input - 3 11 2 4 4 5 6 10 8 -12 Output - 15