Const Qualifier in C - GeeksforGeeks

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Const Qualifier in C

Difficulty Level : Medium
Last Updated : 30 Jul, 2021

The qualifier const can be applied to the declaration of any variable to specify that its value will not be changed ( Which depends upon where const variables are stored, we may change the value of const variable by using pointer ). The result is implementation-defined if an attempt is made to change a const.
1) Pointer to variable.

int *ptr;

We can change the value of ptr and we can also change the value of object ptr pointing to. Pointer and value pointed by pointer both are stored in the read-write area. See the following code fragment.

#include <stdio.h>

int main(void)

{

int i = 10;

int j = 20;

int *ptr = &i;

/* pointer to integer */

printf("*ptr: %d\n", *ptr);

/* pointer is pointing to another variable */

ptr = &j;

printf("*ptr: %d\n", *ptr);

/* we can change value stored by pointer */

*ptr = 100;

printf("*ptr: %d\n", *ptr);

return 0;

}

Output:

*ptr: 10*ptr: 20*ptr: 100

2) Pointer to constant.
Pointer to constant can be declared in following two ways.

const int *ptr;

int const *ptr;

We can change the pointer to point to any other integer variable, but cannot change the value of the object (entity) pointed using pointer ptr. The pointer is stored in the read-write area (stack in the present case). The object pointed may be in the read-only or read-write area. Let us see the following examples.

#include <stdio.h>

int main(void)

{

int i = 10;

int j = 20;

/* ptr is pointer to constant */

const int *ptr = &i;

printf("ptr: %d\n", *ptr);

/* error: object pointed cannot be modified

using the pointer ptr */

*ptr = 100;

ptr = &j; /* valid */

printf("ptr: %d\n", *ptr);

return 0;

}

Output:

error: assignment of read-only location ‘*ptr’

Following is another example where variable i itself is constant.

#include <stdio.h>

int main(void)

{

/* i is stored in read only area*/

int const i = 10;

int j = 20;

/* pointer to integer constant. Here i

is of type "const int", and &i is of

type "const int *". And p is of type

"const int", types are matching no issue */

int const *ptr = &i;

printf("ptr: %d\n", *ptr);

/* error */

*ptr = 100;

/* valid. We call it up qualification. In

C/C++, the type of "int *" is allowed to up

qualify to the type "const int *". The type of

&j is "int *" and is implicitly up qualified by

the compiler to "const int *" */

ptr = &j;

printf("ptr: %d\n", *ptr);

return 0;

}

Output:

error: assignment of read-only location ‘*ptr’

Down qualification is not allowed in C++ and may cause warnings in C. Following is another example with down qualification.

#include <stdio.h>

int main(void)

{

int i = 10;

int const j = 20;

/* ptr is pointing an integer object */

int *ptr = &i;

printf("*ptr: %d\n", *ptr);

/* The below assignment is invalid in C++, results in error

In C, the compiler *may* throw a warning, but casting is

implicitly allowed */

ptr = &j;

/* In C++, it is called 'down qualification'. The type of expression

&j is "const int *" and the type of ptr is "int *". The

assignment "ptr = &j" causes to implicitly remove const-ness

from the expression &j. C++ being more type restrictive, will not

allow implicit down qualification. However, C++ allows implicit

up qualification. The reason being, const qualified identifiers

are bound to be placed in read-only memory (but not always). If

C++ allows above kind of assignment (ptr = &j), we can use 'ptr'

to modify value of j which is in read-only memory. The

consequences are implementation dependent, the program may fail

at runtime. So strict type checking helps clean code. */

printf("*ptr: %d\n", *ptr);

return 0;

}

// Reference:

// http://www.dansaks.com/articles/1999-02%20const%20T%20vs%20T%20const.pdf

// More interesting stuff on C/C++ @ http://www.dansaks.com/articles.shtml

3) Constant pointer to variable.

int *const ptr;

Above declaration is a constant pointer to an integer variable, means we can change the value of object pointed by pointer, but cannot change the pointer to point another variable.

#include <stdio.h>

int main(void)

{

int i = 10;

int j = 20;

/* constant pointer to integer */

int *const ptr = &i;

printf("ptr: %d\n", *ptr);

*ptr = 100; /* valid */

printf("ptr: %d\n", *ptr);

ptr = &j; /* error */

return 0;

}

Output:

error: assignment of read-only variable ‘ptr’

4) constant pointer to constant

const int *const ptr;

Above declaration is a constant pointer to a constant variable which means we cannot change value pointed by the pointer as well as we cannot point the pointer to other variables. Let us see with an example.

#include <stdio.h>

int main(void)

{

int i = 10;

int j = 20;

/* constant pointer to constant integer */

const int *const ptr = &i;

printf("ptr: %d\n", *ptr);

ptr = &j; /* error */

*ptr = 100; /* error */

return 0;

}

Output:

error: assignment of read-only variable ‘ptr’error: assignment of read-only location ‘*ptr’

Summary :

TypeDeclaration

pointer value change

( *ptr = 100 )

pointing value change

( ptr = &a)

1) Pointer to Variableint * ptr yes yes2) Pointer to Constant

const int * ptr
int const * ptr

no yes3) Constant Pointer to Variableint * const ptr yes no4) Constant Pointer to Constant const int * const ptrno no

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