printing array elements

[crazy]

Hi
It seems something really simple, but can't figure out.
It's not my homework :), was just looking at some C problems on the web
and came across this one.
why doesn't the following code print anything?

#include <stdio.h>

#define TOTAL_ELEMENTS (sizeof(array) / sizeof(array[0]))
int array[] = {1,2,3,4,5,6,7} ;

int main()
{
int d;
for(d=-1;d <= (TOTAL_ELEMENTS - 2);d++)
printf("%d\n",a rray[d+1]);
return 0;
}

crazy

[Mike Wahler]

"crazy" <ia*********@ho tmail.com> wrote in message
news:qCIeb.4778 56$cF.162833@rw crnsc53...

Hi
It seems something really simple, but can't figure out.
It's not my homework :), was just looking at some C problems on the web
and came across this one.
why doesn't the following code print anything?

#include <stdio.h>

#define TOTAL_ELEMENTS (sizeof(array) / sizeof(array[0]))
int array[] = {1,2,3,4,5,6,7} ;

int main()
{
int d;
for(d=-1;d <= (TOTAL_ELEMENTS - 2);d++)
for(d=-1;d <= (int)(TOTAL_ELE MENTS - 2);d++)
printf("%d\n",a rray[d+1]);
return 0;
}

Read about 'signed', 'unsigned', and 'conversions'.

IMO it's poor practice to use negative offsets
with arrays like this. Or was this just a 'puzzle'
to be solved?

-Mike

[Mike Wahler]

"Mike Wahler" <mk******@mkwah ler.net> wrote in message
news:0m******** **********@news read3.news.pas. earthlink.net.. .

"crazy" <ia*********@ho tmail.com> wrote in message
news:qCIeb.4778 56$cF.162833@rw crnsc53...

Hi
It seems something really simple, but can't figure out.
It's not my homework :), was just looking at some C problems on the web
and came across this one.
why doesn't the following code print anything?

#include <stdio.h>

#define TOTAL_ELEMENTS (sizeof(array) / sizeof(array[0]))
int array[] = {1,2,3,4,5,6,7} ;

int main()
{
int d;
for(d=-1;d <= (TOTAL_ELEMENTS - 2);d++)

for(d=-1;d <= (int)(TOTAL_ELE MENTS - 2);d++)

printf("%d\n",a rray[d+1]);
return 0;
}

Read about 'signed', 'unsigned', and 'conversions'.

IMO it's poor practice to use negative offsets
with arrays like this. Or was this just a 'puzzle'
to be solved?

Additonal hint: 'sizeof' returns an (impelemenation-
defined) unsigned type.

-Mike

[Andreas Kahari]

In article <qCIeb.477856$c F.162833@rwcrns c53>, crazy wrote:

Hi
It seems something really simple, but can't figure out.
It's not my homework :), was just looking at some C problems on the web
and came across this one.
why doesn't the following code print anything?

#include <stdio.h>

#define TOTAL_ELEMENTS (sizeof(array) / sizeof(array[0]))
int array[] = {1,2,3,4,5,6,7} ;

int main()
{
int d;
for(d=-1;d <= (TOTAL_ELEMENTS - 2);d++)
printf("%d\n",a rray[d+1]);
return 0;
}

Cast "(TOTAL_ELEMENT S - 2)" to int and all will be fine.

To explore further, run this:

#include <stdio.h>
#include <stddef.h>

int
main()
{
size_t i = 10; /* unsigned integer type */
int j = -1; /* signed integer type */

if (j < i) { /* what gets converted to what? */
printf("j < i\n"); /* will not print */
} else {
printf("j >= i\n"); /* will print */
}

return 0;
}

C99 says, "if the operand that has unsigned integer type has
rank greater or equal to the rank of the type of the other
operand, then the operand with signed integer type is converted
to the type of the operand with unsigned integer type."

So, size_t must have the same or greater rank than int. In my
smaller example, the -1 in j gets converted into an unsigned
integer type. Try printing -1 cast to size_t and see what it
is.

--
Andreas Kähäri

[crazy]

"Andreas Kahari" <ak*******@free shell.org> wrote in message
news:sl******** **************@ vinland.freeshe ll.org...

In article <qCIeb.477856$c F.162833@rwcrns c53>, crazy wrote:
----snipped----

Cast "(TOTAL_ELEMENT S - 2)" to int and all will be fine.

To explore further, run this:

#include <stdio.h>
#include <stddef.h>

int
main()
{
size_t i = 10; /* unsigned integer type */
int j = -1; /* signed integer type */

if (j < i) { /* what gets converted to what? */
printf("j < i\n"); /* will not print */
} else {
printf("j >= i\n"); /* will print */
}

return 0;
}

This example definitely helped understand what was wrong.

My mistake was printing d as %d and not as %u.
The moment I did that, it became very clear what was happening.

C99 says, "if the operand that has unsigned integer type has
rank greater or equal to the rank of the type of the other
operand, then the operand with signed integer type is converted
to the type of the operand with unsigned integer type."
What is meant by rank here? what is the rank precedence?

So, size_t must have the same or greater rank than int. In my
smaller example, the -1 in j gets converted into an unsigned
integer type. Try printing -1 cast to size_t and see what it
is.

--
Andreas Kähäri

-crazy

[Peter Nilsson]

"crazy" <ia*********@ho tmail.com> wrote in message news:<qCIeb.477 856$cF.162833@r wcrnsc53>...

why doesn't the following code print anything?

#include <stdio.h>

#define TOTAL_ELEMENTS (sizeof(array) / sizeof(array[0]))
int array[] = {1,2,3,4,5,6,7} ;

int main()
{
int d;
for(d=-1;d <= (TOTAL_ELEMENTS - 2);d++)
printf("%d\n",a rray[d+1]);
return 0;
}

Because size_t has equal or higher rank than an int on your machine.
Try experimenting with (-1 < 1u).

--
Peter

[Jack Klein]

On Thu, 02 Oct 2003 00:02:58 GMT, "crazy" <ia*********@ho tmail.com>
wrote in comp.lang.c:

"Andreas Kahari" <ak*******@free shell.org> wrote in message
news:sl******** **************@ vinland.freeshe ll.org...

In article <qCIeb.477856$c F.162833@rwcrns c53>, crazy wrote:

----snipped----

Cast "(TOTAL_ELEMENT S - 2)" to int and all will be fine.

To explore further, run this:

#include <stdio.h>
#include <stddef.h>

int
main()
{
size_t i = 10; /* unsigned integer type */
int j = -1; /* signed integer type */

if (j < i) { /* what gets converted to what? */
printf("j < i\n"); /* will not print */
} else {
printf("j >= i\n"); /* will print */
}

return 0;
}

This example definitely helped understand what was wrong.

My mistake was printing d as %d and not as %u.
The moment I did that, it became very clear what was happening.

C99 says, "if the operand that has unsigned integer type has
rank greater or equal to the rank of the type of the other
operand, then the operand with signed integer type is converted
to the type of the operand with unsigned integer type."

What is meant by rank here? what is the rank precedence?

So, size_t must have the same or greater rank than int. In my
smaller example, the -1 in j gets converted into an unsigned
integer type. Try printing -1 cast to size_t and see what it
is.

--
Andreas Kähäri

-crazy

The integer types have a ranking in C, such that each higher-ranked
integer type must be able to contain at least the range of values of
lesser ranked types, if not more. This is also subject to the minimum
allowable absolute range for each type.

Signed or unsigned short int must be able to hold all the valuer of
signed and unsigned char, respectively.

(un)signed int must be able to hold all the values of (un)signed
short.

(un)signed long must be able to hold all the values of (un)signed int.

(un)signed long long must be able to hold all the values of (un)signed
long.

--
Jack Klein
Home: http://JK-Technology.Com
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[Dan Pop]

In <SW************ ********@rwcrns c51.ops.asp.att .net> "crazy" <ia*********@ho tmail.com> writes:

My mistake was printing d as %d and not as %u.
Nope, that was NOT your mistake: d has type int, therefore it CANNOT be
printed with %u, which expects an unsigned int!
The moment I did that, it became very clear what was happening.

To do that properly, you must use (unsigned)d.

C99 says, "if the operand that has unsigned integer type has
rank greater or equal to the rank of the type of the other
operand, then the operand with signed integer type is converted
to the type of the operand with unsigned integer type."

What is meant by rank here? what is the rank precedence?

At your current level, you can safely ignore the C99 lingo. Simply
read about the arithmetic conversions in your favourite C book.

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de