Question

#include<stdio.h>
main()
{

        char c = 'R';
        printf("%c\n",c);
        c++;
        printf("%c\n",c);
        char *ptr ="Ramco Systems";
        printf("%c\n",(*ptr));
        (*ptr)++;
        printf("%d\n",(*ptr));

}

The output of the first, second ,3rd printf are ‘R’, ‘S’ & ‘R’ (as expected). However the line “(*ptr)++;” gives runtime error. Can someone explain why ?

Answer 1

The reason is because the memory pointed to be ptr was set at compiletime and is non-modifiable.

So accessing the first character via *ptr is fine and returns R, but attempting to increment the first character yields a runtime error because you are not allowed to modify strings that you provide at compiletime.

To expand on Seg Fault’s comment below, a better way to write your code would have been:

const char *ptr ="Ramco Systems"; //pointer to const char
(*ptr)++; // yields compiletime error because *ptr is a const char

Notice how in this new code, the declared pointer type is more accurate and as a result the compiler is able to give a compiletime error on the second line (much better than a runtime error).