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Accessing void * buffer/array through char * pointer

This code

#include <stdio.h>

int main(void)
{
int hello[] = {'h', 'e', 'l', 'l', 'o'};
char *p = (void *) hello;

for (size_t i = 0; i < sizeof(hello); ++i) {
printf("byte %2zu: <%c>", i, p[i]);
if (!p[i])
printf(" (null char)");
printf("\n");
}

return 0;
}

produces this output

byte 0: <h>
byte 1: <(null char)
byte 2: <(null char)
byte 3: <(null char)
byte 4: <e>
byte 5: <(null char)
byte 6: <(null char)
byte 7: <(null char)
byte 8: <l>
byte 9: <(null char)
byte 10: <(null char)
byte 11: <(null char)
byte 12: <l>
byte 13: <(null char)
byte 14: <(null char)
byte 15: <(null char)
byte 16: <o>
byte 17: <(null char)
byte 18: <(null char)
byte 19: <(null char)

I'm confused about the int *-to-void *-to char * conversion. The
output shows that ints are four bytes on my machine, and the values in
the initializer 'h', 'e', 'l', 'l', 'o' have the values 104, 101, 108,
108, and 111, respectively, so they're able to be represented as chars
(duh...). But if I'd change the initializer to something like

int hello[] = {1000, 43676, 362, 6364, 2575};

I'd get this output

byte 0: <�>
byte 1: <>
byte 2: <(null char)
byte 3: <(null char)
byte 4: <�>
byte 5: <�>
byte 6: <(null char)
byte 7: <(null char)
byte 8: <j>
byte 9: <>
byte 10: <(null char)
byte 11: <(null char)
byte 12: <�>
byte 13: <▒>
byte 14: <(null char)
byte 15: <(null char)
byte 16: <>
byte 17: <
>
byte 18: <(null char)
byte 19: <(null char)

(In other words, non-printable characters.)

Is some kind of overflow happening when I subscript the char pointer?
Or am I simply getting meaningless values because of accessing a char
pointer that points to something that wasn't a char object?

Sebastian

Sep 6 '08 #1
16 6767
s0****@gmail.co m writes:
#include <stdio.h>

int main(void)
{
int hello[] = {'h', 'e', 'l', 'l', 'o'};
char *p = (void *) hello;

for (size_t i = 0; i < sizeof(hello); ++i) {
printf("byte %2zu: <%c>", i, p[i]);
if (!p[i])
printf(" (null char)");
printf("\n");
}

return 0;
}
<snip>
I'm confused about the int *-to-void *-to char * conversion. The
output shows that ints are four bytes on my machine, and the values in
the initializer 'h', 'e', 'l', 'l', 'o' have the values 104, 101, 108,
108, and 111, respectively, so they're able to be represented as chars
(duh...). But if I'd change the initializer to something like

int hello[] = {1000, 43676, 362, 6364, 2575};

I'd get this output

byte 0: <�>
byte 1: <>
byte 2: <(null char)
byte 3: <(null char)
byte 4: <�>
byte 5: <�>
byte 6: <(null char)
byte 7: <(null char)
<snip>
(In other words, non-printable characters.)

Is some kind of overflow happening when I subscript the char
pointer?
No, no overflow is happening on access.
Or am I simply getting meaningless values because of accessing a char
pointer that points to something that wasn't a char object?
First, they are not meaningless. Some process governs exactly what
you see but since it may involve things like the terminal setting it
can be a very complex one.

Secondly, a char pointer always points at a char object. C does not
mandate the value but any object of any type can be accessed as if it
is a sequence of char objects. In your case the first two characters
are almost certainly 1000 % 256 and 1000 / 256, i.e. the least and
second least significant bytes of the binary representation of 1000.

I know this is not an actual answer, but your either/or questions
don't give me much room!

--
Ben.
Sep 6 '08 #2
On Sep 6, 4:26 pm, s0s...@gmail.co m wrote:
This code

#include <stdio.h>

int main(void)
{
int hello[] = {'h', 'e', 'l', 'l', 'o'};
char *p = (void *) hello;

for (size_t i = 0; i < sizeof(hello); ++i) {
printf("byte %2zu: <%c>", i, p[i]);
if (!p[i])
printf(" (null char)");
printf("\n");
}

return 0;

}

produces this output

byte 0: <h>
byte 1: <(null char)
byte 2: <(null char)
byte 3: <(null char)
byte 4: <e>
byte 5: <(null char)
byte 6: <(null char)
byte 7: <(null char)
byte 8: <l>
byte 9: <(null char)
byte 10: <(null char)
byte 11: <(null char)
byte 12: <l>
byte 13: <(null char)
byte 14: <(null char)
byte 15: <(null char)
byte 16: <o>
byte 17: <(null char)
byte 18: <(null char)
byte 19: <(null char)

I'm confused about the int *-to-void *-to char * conversion. The
output shows that ints are four bytes on my machine, and the values in
the initializer 'h', 'e', 'l', 'l', 'o' have the values 104, 101, 108,
108, and 111, respectively, so they're able to be represented as chars
(duh...). But if I'd change the initializer to something like

int hello[] = {1000, 43676, 362, 6364, 2575};

I'd get this output

byte 0: < >
byte 1: <>
byte 2: <(null char)
byte 3: <(null char)
byte 4: < >
byte 5: < >
byte 6: <(null char)
byte 7: <(null char)
byte 8: <j>
byte 9: <>
byte 10: <(null char)
byte 11: <(null char)
byte 12: < >
byte 13: <>
byte 14: <(null char)
byte 15: <(null char)
byte 16: <>
byte 17: <

byte 18: <(null char)
byte 19: <(null char)

(In other words, non-printable characters.)

Is some kind of overflow happening when I subscript the char pointer?
Or am I simply getting meaningless values because of accessing a char
pointer that points to something that wasn't a char object?

Sebastian
your question is same as i asked here before acutally acutally when we
have array of chars or int the linker allocates bytes of initialized
values plus 4 BYTES MORE which also include a NuLL char '\0'
Sep 6 '08 #3
On Sep 6, 4:26 pm, s0s...@gmail.co m wrote:
This code

#include <stdio.h>

int main(void)
{
int hello[] = {'h', 'e', 'l', 'l', 'o'};
char *p = (void *) hello;

for (size_t i = 0; i < sizeof(hello); ++i) {
printf("byte %2zu: <%c>", i, p[i]);
if (!p[i])
printf(" (null char)");
printf("\n");
}

return 0;

}

produces this output

byte 0: <h>
byte 1: <(null char)
byte 2: <(null char)
byte 3: <(null char)
byte 4: <e>
byte 5: <(null char)
byte 6: <(null char)
byte 7: <(null char)
byte 8: <l>
byte 9: <(null char)
byte 10: <(null char)
byte 11: <(null char)
byte 12: <l>
byte 13: <(null char)
byte 14: <(null char)
byte 15: <(null char)
byte 16: <o>
byte 17: <(null char)
byte 18: <(null char)
byte 19: <(null char)

I'm confused about the int *-to-void *-to char * conversion. The
output shows that ints are four bytes on my machine, and the values in
the initializer 'h', 'e', 'l', 'l', 'o' have the values 104, 101, 108,
108, and 111, respectively, so they're able to be represented as chars
(duh...). But if I'd change the initializer to something like

int hello[] = {1000, 43676, 362, 6364, 2575};

I'd get this output

byte 0: < >
byte 1: <>
byte 2: <(null char)
byte 3: <(null char)
byte 4: < >
byte 5: < >
byte 6: <(null char)
byte 7: <(null char)
byte 8: <j>
byte 9: <>
byte 10: <(null char)
byte 11: <(null char)
byte 12: < >
byte 13: <>
byte 14: <(null char)
byte 15: <(null char)
byte 16: <>
byte 17: <

byte 18: <(null char)
byte 19: <(null char)

(In other words, non-printable characters.)

Is some kind of overflow happening when I subscript the char pointer?
Or am I simply getting meaningless values because of accessing a char
pointer that points to something that wasn't a char object?

Sebastian
here is the post
http://groups.google.co.in/group/com...42714e7b3e6c17
Sep 6 '08 #4
On Sep 6, 2:26 pm, s0s...@gmail.co m wrote:
This code
Is broken. :P
#include <stdio.h>

int main(void)
{
int hello[] = {'h', 'e', 'l', 'l', 'o'};
char *p = (void *) hello;
Change p to `unsigned char'. And the cast can be (void *) or (unsigned
char *).
for (size_t i = 0; i < sizeof(hello); ++i) {
printf("byte %2zu: <%c>", i, p[i]);
Remove the 2 in %2zu, else the output might not be meaningful.
Evaluating p[i] can invoke undefined behavior. Changing p to type
`unsigned char *' as I have suggested previously fixes this.
if (!p[i])
printf(" (null char)");
printf("\n");
}

return 0;

}

produces this output
<snip output>

I'm confused about the int *-to-void *-to char * conversion. The
output shows that ints are four bytes on my machine, and the values in
the initializer 'h', 'e', 'l', 'l', 'o' have the values 104, 101, 108,
108, and 111, respectively, so they're able to be represented as chars
(duh...). But if I'd change the initializer to something like
You can convert any pointer to object to unsigned char * to inspect
its object representation.
int hello[] = {1000, 43676, 362, 6364, 2575};

I'd get this output
<snip>
(In other words, non-printable characters.)
So what?
Is some kind of overflow happening when I subscript the char pointer?
Or am I simply getting meaningless values because of accessing a char
pointer that points to something that wasn't a char object?
*ASSUMING* you change p to unsigned char *, you split the objects
representation in CHAR_BIT chunks, and you treat those bits as value
bits, even though in the original object they might be padding bits or
a sign bit.

Change your program to this for more meaningful output:

#include <stdio.h>

typedef int object_type;
#define SIZE 10

int main(void) {

object_type object[SIZE];
unsigned char *p;
size_t i;

p = (unsigned char *)object;

for(i = 0; i < sizeof object; i++)
if(isprint(p[i])) printf("p[%zu] = '%c'\n", i, p[i]);
else printf("p[%zu] = not printable\n", i);

return 0;
}
Sep 6 '08 #5
raashid bhatt said:

<snip>
your question is same as i asked here before acutally acutally when we
have array of chars or int the linker allocates bytes of initialized
values plus 4 BYTES MORE which also include a NuLL char '\0'
You were told you were wrong before, and why. You are wrong now, for the
same reason.

--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999
Sep 6 '08 #6
raashid bhatt <ra**********@g mail.comwrites:
On Sep 6, 4:26 pm, s0s...@gmail.co m wrote:
<snip>
>int main(void)
{
int hello[] = {'h', 'e', 'l', 'l', 'o'};
char *p = (void *) hello;
<snip>
>}
<snip>
here is the post
http://groups.google.co.in/group/com...42714e7b3e6c17
In case you are confused by this reply, there is no connection between
your code and the thread cited by raashid bhatt.

--
Ben.
Sep 6 '08 #7

<s0****@gmail.c omwrote in message
Is some kind of overflow happening when I subscript the char pointer?
Or am I simply getting meaningless values because of accessing a char
pointer that points to something that wasn't a char object?
It's the second. You're using a char pointer to point to a list of integers.
On your machine, as is typical, ints are 4 bytes wide, i.e. 4 chars. So you
are printing out the internal representation of an integer, in its
ASCII-encoded value. However it isn't too meaningless because the low byte
happens to be set to a constant value that is a char, whilst the high bytes
are zero. If your machine was big-endian you'd see <null<null<null h
and so on.

--
Free games and programming goodies.
http://www.personal.leeds.ac.uk/~bgy1mm
Sep 6 '08 #8
vi******@gmail. com writes:
On Sep 6, 2:26 pm, s0s...@gmail.co m wrote:
<snip>
> for (size_t i = 0; i < sizeof(hello); ++i) {
printf("byte %2zu: <%c>", i, p[i]);

Remove the 2 in %2zu, else the output might not be meaningful.
What on earth is wrong with the 2?
Evaluating p[i] can invoke undefined behavior.
I think it is better to say "may invoke undefined behaviour". I
accept that you don't agree (there's been a long thread about this
already) but just for the benefit of the OP there are systems on which
the code posted can't go wrong in any way. Using a potentially signed
char pointer merely limits the portability to a very specific class of
implementations and you, as the programmer, can know (with absolute
certainty) if the code's behaviour is defined or not beforehand. This
is quite unlike some other kinds of UB.

However, one should always used unsigned char for this purpose since
there is no advantage to be gained by using char *.

--
Ben.
Sep 6 '08 #9
On Sep 6, 7:02*am, Ben Bacarisse <be********@bsb .me.ukwrote:
s0****@gmail.co m writes:
<snip>
>Or am I simply getting meaningless values because of accessing a char
pointer that points to something that wasn't a char object?

First, they are not meaningless. *Some process governs exactly what
you see but since it may involve things like the terminal setting it
can be a very complex one.
Well, I meant meaningless in the sense that the values were produced
in an unnatural way. For example, with the second initializer, the
bits that make up for the 1000 in the first element of the int array
are then broken when accessed through the char pointer and everything
becomes a mess! (I got -24 when trying to see the numerical value of
the first element that the char pointer was pointing to.)
Secondly, a char pointer always points at a char object. *C does not
mandate the value but any object of any type can be accessed as if it
is a sequence of char objects. *In your case the first two characters
are almost certainly 1000 % 256 and 1000 / 256, i.e. the least and
second least significant bytes of the binary representation of 1000.
Changing the printf format string to "byte %2zu: <%d>" (i.e., to print
a number instead of a character) yields -24 (which is not 1000 % 256)
for the first character and 3 (which is indeed 1000 / 256) for the
second.
I know this is not an actual answer, but your either/or questions
don't give me much room!
No, it was indeed useful. Anyway, I was just unsure whether accessing
the char pointer in that way was safe. Thanks.

Sebastian

Sep 6 '08 #10

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