On 2019-02-20 21:34 +0100, Patrick Bégou wrote:
> Le 19/02/2019 à 18:34, Liu Hao a écrit :
> > 在 2019/2/19 23:11, Xi Ruoyao 写道:
> > > On 2019-02-19 15:44 +0100, Patrick Bégou wrote:
> > > > Hi,
> > > >
> > > > I need a small theoritical explanation about arrays in C.
> > > These questions are off-topic in gcc-help. It should go to
> > > comp.lang.c or
> > > somewhere.
> > >
> > > > I would like to know why using:
> > > >
> > > > char A[24];
> > > >
> > > > scanf("%s", A); and scanf("%s", &A); does the same thing (as
> > > > it is not
> > > > the same code).
> > > >
> > > > I've checked that printing A or &A with "%p" prints the same
> > > > address....
> > > > but why ?
> > > STFG. The first URL Google suggested for "address of array" is
> > > exactly the
> > > answer of this question:
> > >
> > > https://stackoverflow.com/questions/2528318/how-come-an-arrays-address-is-equal-to-its-value-in-c
> > >
> > Converting both `A` and `&A` to `void *` will yield the same
> > result,
> > because 'an array type describes a contiguously allocated nonempty
> > set
> > of objects with a particular member object type, called the
> > /element
> > type/' [1]. There have to be nothing else; not even padding bytes
> > are
> > allowed. However, as function arguments, the former decays to `char
> > *`
> > while the latter has exact `char (*)[24]`.
> >
> > `%s` requires a corresponding argument be 'a pointer to the
> > initial element of a character array' [2], where `A` satisfies this
> > requirement, so it is perfectly valid there, albeit unsafe.
> >
> > Passing `&A` as the argument corresponding to either `%s` for
> > `scanf()`
> > or `%p` for `printf()` results in undefined behavior because `char
> > (*)[24]` may have a representation from `char *` [3], which GCC
> > warns
> > about if `-Wpedantic` is used with `-Wformat` or `-Wall`.
> >
> > [1] ISO/IEC WG14 Draft N2176, 6.2.5 Types, 20
> > [2] ISO/IEC WG14 Draft N2176, 7.21.6.2 The fscanf function, 12
> > [3] http://c-faq.com/null/machexamp.html
> >
> Thanks a lot Liu, very instructive and detailed explanation.
> I had really a basic understanding of this and had some trouble to
> answer a student question "why does it work with and without the
> &...."
It might not work, as Liu said. scanf/printf always invoke undefined
behavior if the argument corresponding to "%s" is not a pointer to
char. It just happens to work with x86 and x86_64, and most OS, C
libraries, compilers supporting them because all pointers have same
presentation.
In most circumstances you don't need to use unary "&" on an array.
I've never seen it in "real" software code.
The stupid textbook provided by my university even claimed "you can't
use unary & on an array". That kind of bulls**t made me throw it away
and buy a TCPL insteadly. But, OTOH, this also stopped some newbies to
misuse unary &. Maybe a good thing :(.
--
Xi Ruoyao <xry111@xxxxxxxxxxxxxxxx>
School of Aerospace Science and Technology, Xidian University
