Thread: Type Qualifier const

>char const *str;
This causes the object being pointed to to be read-only

>char *const str;
This causes the pointer itself to be read-only. You can also do this to make both the pointer and the object being pointed to read-only:
const char * const str;
or
char const * const str;

>what is the effect of type qualifier const?
The function is passed a copy of the variable, but the compiler will complain if you try to change the copy, for example:

Code:

#include <stdio.h>
#include <string.h>

void fun ( char * const input ) /* Pointer is read-only */
{
  char input2[BUFSIZ] = "World";
  puts ( input );
  /* Attempt to change the address the pointer points to
  ** which will cause an error
  */
  input = input2;
  puts ( input );
}

int main ( void )
{
 char input[BUFSIZ] = "Hello";
 fun ( input );
 return 0;
}

>There's no way to modify its value anyway
Not quite, everything is passed by value in C, but back to the question. You can modify the value that you pass to the function, but it will be a copy of the variable that you passed.

-Prelude

>However, how can I make the object pointed to read-only?
Your code is an example of how const isn't
This is a case where you really should heed the warnings that your compiler gives you, when you use strcpy to transfer the data then you only get warnings but if you transfer each element manually then you'll get errors:

Code:

for ( i = 0; i < len; i++ )
  input[i] = input2[i]; 
  /* Error, Lvalue specifies const object */

The const qualifier can be overridden by using casting and pointers to go around it. Hence, why strcpy only gave you warnings and not errors.

-Prelude