See this example: http://coliru.stacked-crooked.com/a/048b4aa5046da11b
In this example the function is called recursively.
During each call a pointer to that local areay is appended to a static array of
pointers.
Should a new instance of that local array of const int be created every time,
abort() will never be called.
Since calling a library function is observable behavior, clang's optimization
has effectively changed that program's behavior. Hence I think it is wrong.
[code]
#include <stdlib.h>
static const int *ptrs[2];
static unsigned recur;
void foo(){
const int a[] = {0,1,2,3,4,5,6,7,8,9};
ptrs[recur] = a;
if(recur == 0){
++recur;
foo();
}
if(ptrs[0] == ptrs[1]){
abort();
}
}
int main(){
foo();
}
[/code]
------------------
Best regards,
lh_mouse
2016-04-21
-------------------------------------------------------------
发件人：Jonathan Wakely <[email protected]>
发送日期：2016-04-21 01:51
收件人：lh_mouse
抄送：Bingfeng Mei,gcc
主题：Re: Why does gcc generate const local array on stack?
On 20 April 2016 at 18:31, lh_mouse wrote:
> I tend to say clang is wrong here.
If you can't detect the difference then it is a valid transformation.
> Your identifier 'a' has no linkage. Your object designated by 'a' does not
> have a storage-class specifier.
> So it has automatic storage duration and 6.2.4/7 applies: 'If the scope is
> entered recursively, a new instance of the object is created each time.'
How do you tell the difference between a const array that is recreated
each time and one that isn't?
> Interesting enough, ISO C doesn't say whether distinct objects should have
> distinct addresses.
> It is worth noting that this is explicitly forbidden in ISO C++ because
> distinct complete objects shall have distinct addresses:
If the object's address doesn't escape from the function then I can't
think of a way to tell the difference.
