about pointer initialization

[nick]

ex1.char q[]="Nick";
q[0]='D';

ex2.char *q ="Nick";
q[0]='D';

ex1 is correct, but ex2 occurs run time error.

why?

thanks!

[Mike Wahler]

Re: about pointer initialization


"nick" <i141802596@yah oo.com> wrote in message
news:dl6o61$1j4 k$1@justice.its c.cuhk.edu.hk.. .[color=blue]
> ex1.char q[]="Nick";
> q[0]='D';
>
> ex2.char *q ="Nick";
> q[0]='D';
>
> ex1 is correct, but ex2 occurs run time error.
>
> why?[/color]

Because attempting to modify a string literal
produces undefined behavior, and 'run time error'
is one of an infinite number of possible manifestations
of undefined behavior. Moral: Don't Do That.

-Mike

[Andre Kostur]

Re: about pointer initialization

I'm surprised this isn't in the FAQ... so here goes:

nick <i141802596@yah oo.com> wrote in news:dl6o61$1j4 k$1
@justice.itsc.c uhk.edu.hk:
[color=blue]
> ex1.char q[]="Nick";[/color]

q is an array of 5 bytes, into which is copied 'N' 'i' 'c' 'k' '\0'
[color=blue]
> q[0]='D';[/color]

Reassigns the first byte.
[color=blue]
>
> ex2.char *q ="Nick";[/color]

q is a pointer to the string literal "Nick", stored somewhere in memory,
potentially in a read-only memory page. A more "correct" way to write this
line is:

const char * q = "Nick";

C++ accepts the non-const version in order to be more compatable with C.
[color=blue]
> q[0]='D';[/color]

Attempts to change the first byte, which is const. As a result, Undefined
Behaviour (in your case, a run-time error).

[Greg Comeau]

Re: about pointer initialization

In article <dl6o61$1j4k$1@ justice.itsc.cu hk.edu.hk>,
nick <i141802596@yah oo.com> wrote:[color=blue]
>ex1.char q[]="Nick";
>q[0]='D';
>
>ex2.char *q ="Nick";
>q[0]='D';
>
>ex1 is correct, but ex2 occurs run time error.
>
>why?[/color]

See http://www.comeaucomputing.com/techtalk/#stringliteral

http://www.comeaucomputing.com/techtalk/#ptrvsarray

--
Greg Comeau / Celebrating 20 years of Comeauity!
Comeau C/C++ ONLINE ==> http://www.comeaucomputing.com/tryitout
World Class Compilers: Breathtaking C++, Amazing C99, Fabulous C90.
Comeau C/C++ with Dinkumware's Libraries... Have you tried it?

[Peter_Julian]

Re: about pointer initialization


"nick" <i141802596@yah oo.com> wrote in message
news:dl6o61$1j4 k$1@justice.its c.cuhk.edu.hk.. .
| ex1.char q[]="Nick";
| q[0]='D';
|
| ex2.char *q ="Nick";
| q[0]='D';
|
| ex1 is correct, but ex2 occurs run time error.
|
| why?
|
| thanks!

ex1.char is not a type.
ex2.char is not a type.
Who knows, maybe ex1 and ex2 are actually types?
So which would q be, the pointer or the array?

Lets suppose...

char q[] = "Nick";
char *p = &q[0]; // its a single char
*p = 'S';
std::cout << q; // prints "Sick" (note*)

Note: lets not get confused here, a char array is not handled like
arrays of other primitive types or arrays of user-defined types when it
comes to output. You can thank compatibility for that headache. A
std::string is a much better candidate here.

std::string s("I'm a string");
std::string *p_s = &s; // pointer to an entire string, not just a char
<ah, relief>

Are ex1 and ex2 structs or classes? That would be nice to know. If
so...char is reserved.

struct ex1 {
char c;
};

ex1 q[] = {'a', 'b', 'c'}; // treat them as what they are
ex1 *p = &q[0]; // p points to the first element

ex1 *p2 = {'u'}; // error
at which point p2 is undefined, null, zipfa, dangerous, bad karma, UB
(pick one)

An unitialized pointer is nothing, in my humble opinion: its not even a
pointer. Call it a null pointer if you like. To me its a black hole. A
few will argue, and disscuss, and scream, and beg but its still a black
hole here. Adamantly so.

The closer you get to a black hole, and the more often you get anywhere
near it, it's eventually bound to suck you in.