What will be the output of the following C code?
#include <stdio.h>
void main()
{
int k = 4;
int *const p = &k;
int r = 3;
p = &r;
printf("%d", p);
}
a)
Address of k
b)
Address of r
c)
Compile time error
d)
Address of k + address of r
Correct answer is option 'C'. Can you explain this answer?

Question

What will be the output of the following C code?

#include <stdio.h>
void main()
{
int k = 4;
int *const p = &k;
int r = 3;
p = &r;
printf("%d", p);
}

a)
Address of k
b)
Address of r
c)
Compile time error
d)
Address of k + address of r

Correct answer is option 'C'. Can you explain this answer?

Answer

Explanation:
- Constant Pointer:
- In the given code, a constant pointer p is declared with the value of k.
- The pointer p is declared as a constant, meaning that it cannot be reassigned to point to a different memory location.
- Assignment Error:
- The statement p = &r; tries to assign the address of variable r to the constant pointer p.
- This will result in a compile-time error because constant pointers cannot be reassigned after initialization.
- Compile Time Error:
- Due to the attempt to assign a new address to the constant pointer p, the code will not compile successfully.
- The compiler will generate an error indicating that the assignment to a constant pointer is not allowed.
Therefore, the correct output of the code will be a Compile Time Error.

Answer

Since the pointer p is declared to be constant, trying to assign it with a new value results in an error.
Output:
$ cc pgm11.c
pgm11.c: In function ‘main’:
pgm11.c:7: error: assignment of read-only variable ‘p’
pgm11.c:8: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int * const’

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