On Thu, May 16, 2002 at 09:02:25AM +0200, Perceval Anichini wrote:
> When you write
> argv + 1, the compiler will understand : compute the address
> of argv, and add one time the size of the type which is pointed by argv.
> I remember to you that argv[1] = argv + 1. Brackets are only syntactic
> s
On Thu, May 16, 2002 at 06:13:08PM +0930, Tom Cook wrote:
> On 0, Perceval Anichini <[EMAIL PROTECTED]> wrote:
> [snip]
> > > Moreover, argv + sizeof (argv[1]) is equal to argv[4] (as sizeof (char*) =
> > 4) ...
> >
> > > No. If that were so then you could not access the list of arguments
> > >
On 0, Perceval Anichini <[EMAIL PROTECTED]> wrote:
[snip]
> > Moreover, argv + sizeof (argv[1]) is equal to argv[4] (as sizeof (char*) =
> 4) ...
>
> > No. If that were so then you could not access the list of arguments
> > to a main function as argv[0], argv[1], argv[2] etc. The compiler
> > k
>>>execv( argv[1], argv + sizeof( argv[1] ) );
>> Could you explain this? Why would this line be *more* correct than the
precedent?
> The second version makes the pointer arithmetic explicit and has the
> correct type for the second argument. The first involves an implicit
> cast from char*
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