Is this valid?
int a[20];
void *b;
b = (void *)a; // b points to a[0]
b += 5*sizeof(*a); // b points to a[5]
a[5] = 100;
printf( "%d\n" , *((int *)b) ); // prints 100
If so, if a had been a struct, would it still work?
Is there a possibility that the array could contain some padding, so
rather than sizeof, the assignment would be
b += 5*( (void *)(&(a[1])) - (void *)(&(a[0]));
which seems more more complex.
Would any padding be incorporated into sizeof anyway?
Sep 16 '08
160  5674 
On Sep 16, 7:59 pm, s0s...@gmail.co m wrote:
On Sep 16, 11:51 am, raph...@gmail.c om wrote:
On Sep 16, 4:20 pm, vipps...@gmail. com wrote:
b = ((unsigned char *)b) + 5 * sizeof *a;
I see, got to switch to char * so that it can be incremented properly.
No, to unsigned char *. But if your compiler can do it with the void
*, then do it with the void *. It's probably specialized for that kind
of stuff anyway.
You can have the "worst advice of the month" clc award.
You have two options, do it portably, or do it with an extension, with
absolutely no loss in efficiency or size, and you advice to choose the
extension.
On Sep 16, 12:06*pm, vi******@gmail. com wrote:
On Sep 16, 7:59 pm, s0s...@gmail.co m wrote:
On Sep 16, 11:51 am, raph...@gmail.c om wrote:
On Sep 16, 4:20 pm, vipps...@gmail. com wrote:
>b = ((unsigned char *)b) + 5 * sizeof *a;
I see, got to switch to char * so that it can be incremented properly..
No, to unsigned char *. But if your compiler can do it with the void
*, then do it with the void *. It's probably specialized for that kind
of stuff anyway.
You can have the "worst advice of the month" clc award.
You have two options, do it portably, or do it with an extension, with
absolutely no loss in efficiency or size, and you advice to choose the
extension.
Well, perhaps the OP doesn't need portability (or at least not this
kind of portability). Besides, void * seems more natural for this kind
of task.
Sebastian vi******@gmail. com writes:
You have two options, do it portably, or do it with an extension, with
absolutely no loss in efficiency or size, and you advice to choose the
extension.
I admit that, occasionally, I've used pointer arithmetic on void
* in cases where GCC is the only compiler that matters. (Code
inside the Linux kernel is one example.) The nice thing about
doing it that way is that it avoids having to insert additional
casts, which are ugly.
I wouldn't do it in a place where portability matters.
--
"To get the best out of this book, I strongly recommend that you read it."
--Richard Heathfield
Richard<rg****@ gmail.comwrites :
vi******@gmail. com writes:
>On Sep 16, 7:13 pm, sh.vi...@gmail. com wrote:
>>when you do
--char *c; c++; ==c is incremented by 1
--long int *x; x++ ==x is incremented by 4.
Wrong, assuming x was initialized, it would be incremented by 1.
Even if it wasnt initialised it would be incremented by something.
If x isn't initialized, referring to its value invokes undefined
behavior. It's likely, but by no means certain, that the behavior
would be *as if* it were incremented. It's also possible, on some
systems, that x could have a value such that attempting to read it
causes a program crash. (Before you ask, no, I don't have an
example.)
>Here's proof:
Proof of nothing. You are, again, being purposely difficult.
The ++ operator increments its operand by 1, by definition. The
question is, 1 what? In the case of:
long int *x = some_value;
x ++;
it advances it by 1 long int object, i.e., causes it to point to the
next adjacent long int object in memory, assuming that such an object
exists; it can also legally point just past the end of an array.
>long int i[1], *p = i, *q = &i[1];
Note that evaluating &i[1] is ok, and equivalent to i+1, but only
because of a special-case rule; see C99 6.5.3.2p3.
>printf("%d\n ", (int)(q - p));
Will always print 1.
Yes, because of the way pointer subtraction is defined.
And the following:
int main() {
long int i[1], *p = i, *q = &i[1];
printf("%u\n",p ++);
printf("%u\n",p ++);
printf("%u\n", (int)(p - q));
}
The first printf gives me:
3214862936
And the second:
3214862940
Now, that is 4. On my machine.
You're using "%u" to print pointer values. Surely you know that
invokes undefined behavior, and I know of common real-world systems
where it won't work (e.g., where int is 32 bits and pointers are 64
bits). You're also using "%u" to print an int value; if you had
written
printf("%u\n", (unsigned)(p - q));
that wouldn't be no problem. (Actually printing a non-negative int
value using %u is probably ok, due to another special-case rule, but
there's no point in taking advantage of that fact.) Finally, p starts
as a pointer to the first and only element of a 1-element array. You
increment it twice, causing it to point *past* the end of the array.
Yet another instance of undefined behavior that happens to "work" on
your system -- and it wasn't even necessary to demonstrate your point.
You're making unwarranted (though commonly valid) assumptions about
pointer representations and conversions. This program:
#include <stdio.h>
int main(void)
{
long array[10];
long *p = &array[0];
long *q = &array[1];
printf("%d %d %d\n",
(int)sizeof(lon g), (int)(q - p), (int)q - (int)p);
return 0;
}
will typically print "4 1 4" on a system where sizeof(long)==4 . I've
worked on a real-world system where it would print "8 1 1" due to a
rather unusual pointer representation.
And the thing is, all this undefined behavior wasn't even necessary to
demonstrate your point. Here's a portable program (with
implementation-defined but not undefined behavior) that illustrates
what you're trying to talk about :
#include <stdio.h>
#include <assert.h>
int main(void)
{
long array[10];
long *p = &array[0];
long *q = &array[1];
int diff = q - p;
int byte_diff = (char*)q - (char*)p;
printf("p = %p\n", (void*)p);
printf("q = %p\n", (void*)q);
printf("sizeof( long) = %d\n", (int)sizeof(lon g));
printf("diff = %d\n", diff);
printf("byte_di ff = %d\n", byte_diff);
assert(diff == 1);
assert(byte_dif f == sizeof(long));
return 0;
}
On my system, I get:
p = 0xbfce19ac
q = 0xbfce19b0
sizeof(long) = 4
diff = 1
byte_diff = 4
And here's the point:
p++ increments p by 1. If p is a long*, this means that it advances
the memory location to which it points by 1 long, or by sizeof(long)
bytes.
--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"
Richard wrote:
vi******@gmail. com writes:
On Sep 16, 7:13 pm, sh.vi...@gmail. com wrote:
when you do
--char *c; c++; ==c is incremented by 1
--long int *x; x++ ==x is incremented by 4.
Wrong, assuming x was initialized, it would be incremented by 1.
Even if it wasnt initialised it would be incremented by something.
On real, widely used machines, attempting to access an unitialized
pointer value can cause your program to abort before that value ever
gets a chance to be incremented.
....
long int i[1], *p = i, *q = &i[1];
printf("%d\n", (int)(q - p));
Will always print 1.
And the following:
int main() {
long int i[1], *p = i, *q = &i[1];
printf("%u\n",p ++);
printf("%u\n",p ++);
That is undefined behavior by reason of trying to print an pointer
value using a format string that calls for an unsigned integer. The
behavior is technically meaningless, though in practice it will
produce much the same effect as casting the the pointer value to
unsigned int on many systems. You could have demonstrated your point
without undefined behavior by using uintptr_t; why didn't you?
printf("%u\n", (int)(p - q));
}
The first printf gives me:
3214862936
And the second:
3214862940
Now, that is 4. On my machine.
Yes, but that difference has no portable meaning, and the
implementation-specific meaning on your machine is (probably) only
that p and q point at locations 4 bytes apart; that doesn't say
anything about how much has been added to q to get the current value
of p.
It is only the third printf() that provides a meaningful answer to the
question of how big the difference is between p and q is. You'll get a
different answer, if you cast both pointer to char* before subtracting
them. On your machine that answer will probably be 4, but that answer
is to a different question, ra*****@gmail.c om writes:
On Sep 16, 4:20 pm, vipps...@gmail. com wrote:
>b = ((unsigned char *)b) + 5 * sizeof *a;
I see, got to switch to char * so that it can be incremented properly.
>Yes, but by using the offsetof() macro in <stddef.h>
I meant to point to the structure.
struct st { int val; } a[100];
void *b;
a[5].val = 100;
b = a;
b = ((unsigned char *)b) + 5 * sizeof *a;
printf( "%d\n" , b->val );
If you want to point to a structure, why not just use a
pointer-to-structure rather than a pointer-to-void?
But if you must use void* for some reason, I think this is more
straightforward :
b = (struct st*)b + 5;
--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"
On Sep 16, 9:46*pm, Keith Thompson <ks...@mib.orgw rote:
sh.vi...@gmail. com writes:
On Sep 16, 8:15*pm, raph...@gmail.c om wrote:
Is this valid?
*In my opinion this
extension is a bad idea; it can be convenient, but it doesn't give you
anything you can't do by other means, it has some bizarre
consequences,
I think Bus Error,on some machines, is one of those consequences.
Thanks all for gcc info. I thought gcc didn't take anything as stride
value for void pointers
--
vIpIn
Keith Thompson <ks***@mib.orgw rites:
Richard<rg****@ gmail.comwrites :
>vi******@gmail. com writes:
>>On Sep 16, 7:13 pm, sh.vi...@gmail. com wrote:
when you do
--char *c; c++; ==c is incremented by 1
--long int *x; x++ ==x is incremented by 4.
Wrong, assuming x was initialized, it would be incremented by 1.
Even if it wasnt initialised it would be incremented by something.
If x isn't initialized, referring to its value invokes undefined
behavior. It's likely, but by no means certain, that the behavior
would be *as if* it were incremented. It's also possible, on some
systems, that x could have a value such that attempting to read it
causes a program crash. (Before you ask, no, I don't have an
example.)
Whatever. x will be incremented.
>
>>Here's proof:
Proof of nothing. You are, again, being purposely difficult.
The ++ operator increments its operand by 1, by definition. The
question is, 1 what? In the case of:
long int *x = some_value;
x ++;
it advances it by 1 long int object, i.e., causes it to point to the
next adjacent long int object in memory, assuming that such an object
exists; it can also legally point just past the end of an array.
>>long int i[1], *p = i, *q = &i[1];
Note that evaluating &i[1] is ok, and equivalent to i+1, but only
because of a special-case rule; see C99 6.5.3.2p3.
>>printf("%d\n" , (int)(q - p));
Will always print 1.
Yes, because of the way pointer subtraction is defined.
>And the following:
int main() {
long int i[1], *p = i, *q = &i[1];
printf("%u\n",p ++);
printf("%u\n",p ++);
printf("%u\n", (int)(p - q));
}
The first printf gives me:
3214862936
And the second:
3214862940
Now, that is 4. On my machine.
You're using "%u" to print pointer values. Surely you know that
invokes undefined behavior, and I know of common real-world systems
Garbage. On my machine it prints a 32 bit value. Pointers are values
which I can printf and see and they correspond to physical memory
locations. Its a number get over it. ja*********@ver izon.net writes:
Richard wrote:
>vi******@gmail. com writes:
On Sep 16, 7:13 pm, sh.vi...@gmail. com wrote:
when you do
--char *c; c++; ==c is incremented by 1
--long int *x; x++ ==x is incremented by 4.
Wrong, assuming x was initialized, it would be incremented by 1.
Even if it wasnt initialised it would be incremented by something.
On real, widely used machines, attempting to access an unitialized
pointer value can cause your program to abort before that value ever
gets a chance to be incremented.
Never come across it. Which machines? I do believe you btw. However in
the great majority (99.9999%= of machines on this planet it will indeed
by incremented.
>
...
long int i[1], *p = i, *q = &i[1];
printf("%d\n", (int)(q - p));
Will always print 1.
And the following:
int main() {
long int i[1], *p = i, *q = &i[1];
printf("%u\n",p ++);
printf("%u\n",p ++);
That is undefined behavior by reason of trying to print an pointer
value using a format string that calls for an unsigned integer. The
behavior is technically meaningless, though in practice it will
produce much the same effect as casting the the pointer value to
unsigned int on many systems. You could have demonstrated your point
without undefined behavior by using uintptr_t; why didn't you?
> printf("%u\n", (int)(p - q));
}
The first printf gives me:
3214862936
And the second:
3214862940
Now, that is 4. On my machine.
Yes, but that difference has no portable meaning, and the
It has a meaning. The VALUE of p is incremented by 4. On my machine. And
probably the OPs.
implementation-specific meaning on your machine is (probably) only
that p and q point at locations 4 bytes apart; that doesn't say
anything about how much has been added to q to get the current value
of p.
It is only the third printf() that provides a meaningful answer to the
question of how big the difference is between p and q is. You'll get a
different answer, if you cast both pointer to char* before subtracting
them. On your machine that answer will probably be 4, but that answer
is to a different question,
-- s0****@gmail.co m wrote:
On Sep 16, 12:06 pm, vi******@gmail. com wrote:
>On Sep 16, 7:59 pm, s0s...@gmail.co m wrote:
>>On Sep 16, 11:51 am, raph...@gmail.c om wrote:
On Sep 16, 4:20 pm, vipps...@gmail. com wrote:
b = ((unsigned char *)b) + 5 * sizeof *a;
I see, got to switch to char * so that it can be incremented properly.
No, to unsigned char *. But if your compiler can do it with the void
*, then do it with the void *. It's probably specialized for that kind
of stuff anyway.
You can have the "worst advice of the month" clc award.
You have two options, do it portably, or do it with an extension, with
absolutely no loss in efficiency or size, and you advice to choose the
extension.
Well, perhaps the OP doesn't need portability (or at least not this
kind of portability). Besides, void * seems more natural for this kind
of task.
"This kind of task" was an artificial bit of code specifically
written to experiment with manipulating pointers. An investigative
doodle, nothing more.
Personally, I'm not as ready as vippstar is to give you the
award; we're only halfway through the month. But I'd be surprised
if you weren't on the ballot a couple weeks from now. Have you
chosen your running mate yet?
-- Er*********@sun .com This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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