On 27 Jan 2009, at 23:07, Graham Cox wrote:
On 28 Jan 2009, at 2:24 am, Jeremy Pereira wrote:
Yes. That is correct, but since buffer is already a pointer to the
first byte of the array and then you are taking a reference to it,
key will end up containing the address of the buffer. You really
need:
uint key = *(uint*)buffer;
That's incorrect. Or rather, in this case it doesn't make any
difference, but the use of '&' is more general as it works whether
buffer is an array or not.
When you declare:
uint8 buffer[8];
You've reserved 8 bytes on the stack. Whether you use 'buffer' or
'&buffer' you obtain the same address, as C treats array variables
as pointers. But:
Having tried it just now, I stand corrected.
However, I would argue that this is the C compiler behaving in a
deliberately inconsistent way, since in C, arrays and pointers are
supposed to be the same thing. As buffer is actually a constant,
taking its address should really result in a compiler error in the
same way as int a = &123;
uint8 buffer;
You get very different results from 'buffer' and '&buffer' so in the
interests of defensive programming, using '&' is allowing your code
to tolerate a change to the buffer declaration without breaking.
Because of the additional cast *(int*) such a change would go
unnoticed by the compiler but probably have a very bad outcome at
runtime. It's a good habit IMO to always take the address in these
cases to make it clear in your code what your intentions were when
you wrote it.
I disagree. Consider the following program:
---
#include <stdio.h>
int buf[2];
static void foo(int buf1[])
{
printf ("buf1 %08x %08x\n", buf1, &buf1);
}
int main ()
{
unsigned int key = *(unsigned int*)&buf;
printf ("buf %08x %08x\n", buf, &buf);
foo(buf);
return 0 ;
}
---
If I compile and run it, I get the following output:
buf 00002030 00002030
buf1 00002030 bffffa00
Clearly if I naively copy and paste
unsigned int key = *(unsigned int*)&buf;
from main to foo and just change the name, I will get the wrong result.
--Graham
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