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 &...."
