Question

Ok, I know that this is invalid

 char char_A = 'A';
    const char * myPtr = &char_A;
    *myPtr = 'J'; // error - can't change value of *myP

[Because we declared a pointer to a constant character]

Why is this valid?

 const char  *linuxDistro[6]={ "Debian", "Ubuntu", "OpenSuse", "Fedora", "Linux Mint", "Mandriva"};

for ( int i=0; i < 6; i++) 
cout << *(linuxDistro+i)<< endl;

*linuxDistro="WhyCanIchangeThis";// should result in an error but doesnt ? 
for ( int i=0; i < 6; i++) 
cout << *(linuxDistro+i)<< endl;

Thanks for looking!

Answer 1

You write

*linuxDistro =  "WhyCanIchangeThis";

which is perfectly valid, because the declaration of linuxDistro was

const char *linuxDistro[6];

i. e. it’s an array of 6 pointers to const char. That is, you can change the pointers themselves, just not the characters pointed to by those pointers. I. e., you can not compile

*linuxDistro[0] = 'B';

to obtain the string "Bebian", becuse the strings contain constant characters…

What you probably want is an array of constant pointers to constant characters:

const char *const linuxDistro[6];