From:	"-hs-" <email DOT invalid AT server DOT invalid>
Newsgroups:	comp.lang.c,comp.os.msdos.djgpp
Subject:	Re: what's this mean?
Date:	Thu, 30 Nov 2000 21:34:43 +0100
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Andrew Clark a écrit dans le message <3a267c5f DOT 0 AT news DOT syr DOT edu>...
>bige.c:15: warning: passing arg 2 of 'mem_err' dicards qualifiers from
>pointer target type
>
>line 15:
>
>mem_err("could not create %s\n", names_east[i]);
>
>where mem_err is a function taking 2 char *'s and returning void.


In other words :

void mem_err(char *, char *)

>names_east is a static const char *[].


>what does this warning mean?


It means that you are playing with fire.


The mem_err() function is designed to accept read-write parameters. It means
that the parameters *must* be read-write.

You have two risky business here:

static const char *names_east[]=
{
   "azer",
   "azer",
   "azer",
};

mem_err("could not create %s\n", names_east[i]);

1 - You put a string literal to a char*.
2 - You put a constant string literal to a char*.

As the function is allowed to change the values of the strings, anything can
happen.

If you want a function to accept pointer to const, you must declare it's
parameter pointer to const:

void mem_err(char const *, char const *)

Now see wether the function can be compiled or not. If not, it means that it
attempt to modify a constant data.

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