Accessing void * buffer/array through char * pointer

[s0suk3]

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);
if (!p)
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: <â–’>
byte 14: <> (null char)
byte 15: <> (null char)
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

 

[Ben Bacarisse]

#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);
if (!p)
printf(" (null char)");
printf("\n");
}

return 0;
}

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)

(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!

 

[raashid bhatt]

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);
if (!p)
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'

 

[raashid bhatt]

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);
if (!p)
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/co...b969f/8642714e7b3e6c17?hl=en#8642714e7b3e6c17

 

[vippstar]

This code

Is broken.

#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);

Remove the 2 in %2zu, else the output might not be meaningful.
Evaluating p can invoke undefined behavior. Changing p to type
`unsigned char *' as I have suggested previously fixes this.

if (!p)
printf(" (null char)");
printf("\n");
}

return 0;

}

produces this 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

(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)) printf("p[%zu] = '%c'\n", i, p);
else printf("p[%zu] = not printable\n", i);

return 0;
}

 

[Ben Bacarisse]

On Sep 6, 4:26 pm, (e-mail address removed) wrote:

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

}

here is the post
http://groups.google.co.in/group/co...b969f/8642714e7b3e6c17?hl=en#8642714e7b3e6c17

In case you are confused by this reply, there is no connection between
your code and the thread cited by raashid bhatt.

 

[Ben Bacarisse]

On Sep 6, 2:26 pm, (e-mail address removed) wrote:

for (size_t i = 0; i < sizeof(hello); ++i) {
printf("byte %2zu: <%c>", i, p);

Remove the 2 in %2zu, else the output might not be meaningful.

What on earth is wrong with the 2?

Evaluating p 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 *.

 

[s0suk3]

(e-mail address removed) writes:

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

 

[vippstar]

On Sep 6, 2:26 pm, (e-mail address removed) wrote:

for (size_t i = 0; i < sizeof(hello); ++i) {
printf("byte %2zu: <%c>", i, p);

Remove the 2 in %2zu, else the output might not be meaningful.

What on earth is wrong with the 2?

Nothing, I got confused with the meaning of `2' in scanf. (I thought
only the first two digits would be printed, like only the first two
digits are matched in scanf)
I apologize for this.

 

[Ben Bacarisse]

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.

OK, fair cop. 1000 % 256 = 232 = 256-24. I.e. -24 is 1000 % 256 when
that bit pattern is interpreted as signed using the most common
representation of signed values (2s complement). Also -24 = 232 (mod
256) i.e. -24 is a value of 1000 mod 256 so the result is
understandable in terms of div and mod. I did not know your char was
signed (though in true that is common).

No, it was indeed useful. Anyway, I was just unsure whether accessing
the char pointer in that way was safe. Thanks.

Now that is another question altogether. It is safe unless your
implementation has signed chars and has trap representations for some
bit patters in signed char (-0 being the most likely on non 2s
complement machines).

In one sense (maximum portability) using char is not safe, but since
any given implementation must tell you if it is not safe, I feel it is
a bit broad to say "undefined". However (and this is a big however)
since there is nothing at all to be gained at all from using char
(rather than unsigned char) you should always use unsigned char when
inspecting the representation of an object.

 

[Ralf Damaschke]

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.

Yes, 1000 % 256 == 232. So your result shows that your characters
are 8 bit wide and signed; and that you use an inappropriate type
to access them when you prefer only non-negative values.

-- Ralf

 

[s0suk3]

Now that is another question altogether.  It is safe unless your
implementation has signed chars and has trap representations for some
bit patters in signed char (-0 being the most likely on non 2s
complement machines).

In one sense (maximum portability) using char is not safe, but since
any given implementation must tell you if it is not safe, I feel it is
a bit broad to say "undefined".  However (and this is a big however)
since there is nothing at all to be gained at all from using char
(rather than unsigned char) you should always use unsigned char when
inspecting the representation of an object.

What is the advantage of using unsigned char over char? Why can it
even invoke undefined behavior as vippstar said? I read somewhere
that, if char happens to be signed, and if the variable will be used
to hold 8-bit characters (i.e., a character whose value requires the
use of the leftmost bit), it was better to use unsigned char if the
variable would ever be used as an integer (otherwise we might end up
with a negative value). Are there any other advantages other than
that? And what's the deal with UB?

Sebastian

 

[Ben Bacarisse]

What is the advantage of using unsigned char over char?

Accessing an arbitrary bit pattern using unsigned char * is 100% safe
on all conforming C implementations, so that is what all wise C
programmers do. There can be no reason no to (that I can see).

This means that discussing what happens when char is used is an
academic exercise. I am all for that, but don't confuse the fact that
problems when using char are rare with any endorsement of using char
rather unsigned char!

Why can it
even invoke undefined behavior as vippstar said?

As far as I can see, using signed char * is only a problem when some
bit patterns are trap representations. This is very rare and I'd be
prepared to bet that you would know if you are using such an
implementation. If you are porting to such an implementation you will
hope that the compiler offers a flag to make char unsigned by default
because quite a lot of old C code using char * to do what you are
doing. This is the UB that vippstar was talking about (I think).

The most likely case being accessing a negative zero on a system where
negative zero is a trap representation rather than an alternative form
of zero (this is specifically permitted). On systems where negative
zero is not a trap, your code won't be undefined but you will be
unable to distinguish between the two forms of zero. I.e. two
different ints might compare the same if you compare them as arrays of
signed char values[1].

I read somewhere
that, if char happens to be signed, and if the variable will be used
to hold 8-bit characters (i.e., a character whose value requires the
use of the leftmost bit), it was better to use unsigned char if the
variable would ever be used as an integer (otherwise we might end up
with a negative value).

When char is signed you can be led into making other mistakes -- for
example forgetting to convert to unsigned before calling one of the
is* functions from ctype.h, or using the char to index an array
without checking for negative values, but these are secondary problems
and are unrelated to what vippstar was referring to.

[1] memcmp never does this.

 

[Harald van Dijk]

(e-mail address removed) said:

What is the advantage of using unsigned char over char?

It's guaranteed that you can treat any object as if it were an array of
unsigned char (subject to syntax rules, of course) without breaking
anything - for example,
[snip]
The same guarantee is not extended to char or signed char.

See 6.2.6.1(3) of C99.

You're correct that this talks of unsigned char only, but:

6.3.2.3p7:
"When a pointer to an object is converted to a pointer to a character
type, the result points to the lowest addressed byte of the object.
Successive increments of the result, up to the size of the object, yield
pointers to the remaining bytes of the object."

6.5p7:
"An object shall have its stored value accessed only by an lvalue
expression that has one of the following types:
[snip]
-- a character type."

6.2.6.1p5:
"Certain object representations need not represent a value of the object
type. If the stored value of an object has such a representation and is
read by an lvalue expression that does not have character type, the
behavior is undefined. If such a representation is produced by a side
effect that modifies all or any part of the object by an lvalue
expression that does not have character type, the behavior is
undefined. Such a representation is called a trap representation."

all talk of "character type", not "unsigned char". So... signed char * can
be used to point to the bytes of an object. It can be used to access those
bytes of the object. And if the byte happens to be a trap representation
for signed char, then the behaviour is still not automatically undefined.
All that 6.2.6.1p3 says is that the values you get this way are not called
the object representation, as far as I can tell.