Subj : Re: malloc()
To   : borland.public.cpp.borlandcpp
From : Sebastian Ledesma
Date : Thu Jun 24 2004 05:54 pm

Wayne:

According to BC help:
"When the operand is of type char (signed or unsigned), sizeof gives the
result 1. When the operand is a non-parameter of array type, the result is
the total number of bytes in the array (in other words, an array name is not
converted to a pointer type). The number of elements in an array equals
sizeof array/ sizeof array[0] ."

but probably it refers to 'fixed size' array (eg: char name[20];) and not to
dinamically created ones (char *name):  too bad.

Saludos
Sebastian.

PS: thanks for the fix.

"Wayne A. King" <waking@idirect.com> escribió en el mensaje
news:40db23c1.18290079@newsgroups.borland.com...
> On Wed, 23 Jun 2004 13:03:35 -0300, "Sebastian Ledesma"
> <labo(no_sp@m)solidyne1.com> wrote:
>
> >Can you try something like this?
>
> I hope not. ;-)
>
> >    char *p=realloc ( strlen( szInput ) + 1 );
>
> Look at the prototype for realloc. It needs two arguments.
> e.g. - char *p=realloc (txtBuf,  strlen( szInput ) + 1 );
>
> >        n=sizeof(txtBuf )-1; //assuming 1 char wide :-)
>
> Nope. The expression sizeof(txtBuf) will give the size of the
> txtBuf pointer variable itself, not the size of the allocation to
> which it points. You can't get the allocated size using the
> sizeof operator. (Which is usually evaluated at compile time
> not at run time.)
>
>
> --
> Wayne A. King
> (waking@idirect.com, Wayne_A_King@compuserve.com)

.