On Tue, 2005-11-22 at 13:43 +1100, Robert Collins wrote:
> On Tue, 2005-11-22 at 13:26 +1100, ashley maher wrote:
> > G'day,
> >
> > I know this is not even C 101 level but would some kind soul please
> > explain to me why this is not even close to working.
> >
> > Regards,
> >
> > Ashley
> >
> > #include <stdio.h>
> > #include <stdlib.h>
> > #include <string.h>
> >
> > int somefunction(char *string1)
> > {
> > char *string2 = "some words\0";
> >
> > string1 = (char *)calloc(strlen(string2 + 1), sizeof (char));
> >
> > strcpy(string1, string2);
> >
> > return 0;
> > }
> >
> >
> > int main ()
> > {
> > char *string;
> >
> > somefunction(string);
> >
> > printf ("\n\nString is: %s\n\n", string);
> >
> > free (string);
> >
> > return 0;
> > }
>
> You are pasing a pointer to char to somefunction. Its not altering the
> referenced memory area, rather its just allocating a new variable
> locally.
>
> int somefunction(char **string1location)
> {
> char * string1 = (char *)calloc(strlen(string2 + 1), sizeof (char));
> *string1location = string1;
> ..
> }
>
> should work better.
>
> Rob
>
thanks Rob
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