[ale] C Programming Question

[JB Wells]

yup my bad. or to be more precise

char **a=(char **)malloc(sizeof(char *) * WHATEVER));

although pointer size is the same ;-)


--- "Eric Z. Ayers" <eric.ayers at mindspring.com> wrote:
> JB Wells wrote:
> > 
> > I think this will work:
> > 
> > char **a=(char **)malloc (sizeof(char **));
> 
> ^^ You need to change this line:
> char **a=(char **)malloc (sizeof(char **) *
> WHATEVER);
> 
> > 
> > for (i=0; i<WHATEVER; i++)
> > {
> >     /* if the pointer will be pointing to a
> string,
> >        you should change it to
> >        (char *)malloc(sizeof(char) *
> strlen(STRING))*/
> > 
> >     a[i]=(char *)malloc(sizeof(char *));
> > }
> > 
> > =====
> > --------------------------
> > John B. Wells IV
> > Application Developer
> > Acterna
> > 6100 Lake Forrest Drive
> > Atlanta, GA  30328
> > 404.531.8938
> > 
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