Array Initialization

[Henry]

I finally thought I had an understanding of multi dimensional arrays in C
when I get this:

#include <stdio.h>

#define max_x 3
#define max_y 5

int array[max_x][max_y];

main()
{
int x,y;

for(x = 0; x < max_x; x++)
for(y=0; y < max_y; y++)
array[x][y] = x * 10 + y;

for(y = 0; y < max_y; y++) {
(void)printf("a rray[%d] ", y);
for(x = 0; x < max_x; x++)
(void)printf("% d ", array[x,y]);
(void)printf("\ n");
}

return 0;
}

This program initializes to :
array[0] 56164 56164 56164
array[1] 56184 56184 56184
array[2] 56204 56204 56204
array[3] 56224 56224 56224
array[4] 56244 56244 56244

Could someone please help me figure out how this program arrives at such a
large number?

[Bigdakine]

>Subject: Array Initialization

From: "Henry" he*****@knology .net
Date: 9/2/03 3:27 PM Hawaiian Standard Time
Message-id: <vl************ @corp.supernews .com>

I finally thought I had an understanding of multi dimensional arrays in C
when I get this:

#include <stdio.h>

#define max_x 3
#define max_y 5

int array[max_x][max_y];

main()
{
int x,y;

for(x = 0; x < max_x; x++)
for(y=0; y < max_y; y++)
array[x][y] = x * 10 + y;

for(y = 0; y < max_y; y++) {
(void)printf("a rray[%d] ", y);
for(x = 0; x < max_x; x++)
(void)printf("% d ", array[x,y]);
(void)printf("\ n");
}

return 0;
}

This program initializes to :
array[0] 56164 56164 56164
array[1] 56184 56184 56184
array[2] 56204 56204 56204
array[3] 56224 56224 56224
array[4] 56244 56244 56244

Could someone please help me figure out how this program arrives at such a
large number?

Heck, I'm interested to know how this even compiled.

(void)printf("% d ", array[x,y]);

Should be .. array[x][y]....

Fortran dies hard, don't it? :-)

Stuart
Dr. Stuart A. Weinstein
Ewa Beach Institute of Tectonics
"To err is human, but to really foul things up
requires a creationist"

[Nick Austin]

On Tue, 2 Sep 2003 21:27:36 -0400, "Henry" <he*****@knolog y.net>
wrote:

I finally thought I had an understanding of multi dimensional arrays in C
when I get this:

#include <stdio.h>

#define max_x 3
#define max_y 5

int array[max_x][max_y];

main()
{
int x,y;

for(x = 0; x < max_x; x++)
for(y=0; y < max_y; y++)
array[x][y] = x * 10 + y;

for(y = 0; y < max_y; y++) {
(void)printf("a rray[%d] ", y);
for(x = 0; x < max_x; x++)
(void)printf("% d ", array[x,y]);
(void)printf("% d ", array[x][y]);

array[x,y] uses the comma operator and is the same as array[y].

You're invoking UB because array[y] has type 'int *' and
you're printing it with %d.
(void)printf("\ n");
}

return 0;
}

This program initializes to :
array[0] 56164 56164 56164
array[1] 56184 56184 56184
array[2] 56204 56204 56204
array[3] 56224 56224 56224
array[4] 56244 56244 56244

Could someone please help me figure out how this program arrives at such a
large number?

Nick.

[Brett Frankenberger]

In article <vl************ @corp.supernews .com>,
Henry <he*****@knolog y.net> wrote:

I finally thought I had an understanding of multi dimensional arrays in C
when I get this:

int array[max_x][max_y];

(void)printf("% d ", array[x,y]);

array[x,y] is not the same thing as array[x][y].
The expression:
x,y
is a comma expression whose value is the value of the second
expression (in this case, y). The first expression (in this case, x)
would be evaluated for its side effects; but in this case, there are
none. So x,y is equivalent to just:
y

So you effectively have
printf("%d ", array[y])

array[y] is type (pointer to int). "%d" requires an int. So you're
using an "int" format specifier to print a (pointer to int); the result
is undefined. Change the code to:
printf("%d ", array[x][y]);
and you'll probably get what you expected.

By the way, I'm curious what platform you are running on. (i.e. what
compiler and what OS and what processor)? I'm somewhat surprised at
the numbers you got. (It's all undefined, of course ... but on a
typical PC platform, I'd have expected the actual result to be much
larger than the numbers you got.)

-- Brett

[Brett Frankenberger]

In article <20************ *************** @mb-m23.aol.com>,
Bigdakine <bi*******@aol. comGetaGrip> wrote:

Heck, I'm interested to know how this even compiled.
(void)printf("% d ", array[x,y]);

Why wouldn't it? 'x,y' is a perfectly valid expression (equivalent to
'y'), and array[y] doesn't violate any constraints (and would even be
useful, defined behavior under some circumstances), so one wouldn't
expect compilation to fail. Of course, array[y] is type (pointer to
int), so attempting to print it with "%d" is undefined behavior.

-- Brett

[Henry]

> is a comma expression whose value is the value of the second

expression (in this case, y). The first expression (in this case, x)
would be evaluated for its side effects; but in this case, there are
none. So x,y is equivalent to just:
y

So you effectively have
printf("%d ", array[y])

Ok thanks, when I changed the way it was typed it worked much better..
Strangly the array[x,y] came straight from the "Practical C" book.

The new initialization is this:
array[0] 0 10 20
array[1] 1 11 21
array[2] 2 12 22
array[3] 3 13 23
array[4] 4 14 24

This seems like a much more plausible answer.. Although I am still a little
confused as to how it arrives at the numbers in that order. Could someone
explain it to me a little clearer? The book is not doing a very good job.

Thanks

[Martin Ambuhl]

Henry wrote:

I finally thought I had an understanding of multi dimensional arrays in C
when I get this:

#include <stdio.h>

#define max_x 3
#define max_y 5

int array[max_x][max_y];

main()
For goodness' sake! Why would anyone use an implicit int as the return
value for main, and then decorate printf calls with casts?
[...]
(void)printf("a rray[%d] ", y);

If you do this to satisfy some version of lint, get one that knows that not
only is an explicit 'int' on main a good idea, but it is required by the
current C standard.

--
Martin Ambuhl

[Irrwahn Grausewitz]

"Henry" <he*****@knolog y.net> wrote in
<vl************ @corp.supernews .com>:

I finally thought I had an understanding of multi dimensional arrays in C
when I get this:

#include <stdio.h>

#define max_x 3
#define max_y 5

int array[max_x][max_y];

main() int main(void){
int x,y;

for(x = 0; x < max_x; x++)
for(y=0; y < max_y; y++)
array[x][y] = x * 10 + y;

for(y = 0; y < max_y; y++) {
(void)printf("a rray[%d] ", y); Get rid of these (void) casts - you don't need them.
for(x = 0; x < max_x; x++)
(void)printf("% d ", array[x,y]); ^^^^^
Shouldn't this read: array[x][y] ???
(void)printf("\ n");
}

return 0;
}

This program initializes to :
array[0] 56164 56164 56164
array[1] 56184 56184 56184
array[2] 56204 56204 56204
array[3] 56224 56224 56224
array[4] 56244 56244 56244
Not at all - the initialization is just perfect.
Could someone please help me figure out how this program arrives at such a
large number?

You print array[x,y], which is equivalent to array[y], which is
in turn a pointer to an array of five ints, which you print using
the %d format specifier - hence the 'strange' values.

Irrwahn
--
Rain is just liquid sunshine.

[A. Bolmarcich]

In article <vl************ @corp.supernews .com>, Henry wrote:

I finally thought I had an understanding of multi dimensional arrays in C
when I get this:

#include <stdio.h>

#define max_x 3
#define max_y 5

int array[max_x][max_y];

main()
{
int x,y;

for(x = 0; x < max_x; x++)
for(y=0; y < max_y; y++)
array[x][y] = x * 10 + y;

for(y = 0; y < max_y; y++) {
(void)printf("a rray[%d] ", y);
for(x = 0; x < max_x; x++)
(void)printf("% d ", array[x,y]);
(void)printf("\ n");
}

return 0;
}

This program initializes to :
array[0] 56164 56164 56164
array[1] 56184 56184 56184
array[2] 56204 56204 56204
array[3] 56224 56224 56224
array[4] 56244 56244 56244

Could someone please help me figure out how this program arrives at such a
large number?

The statement

(void)printf("% d ", array[x,y]);

prints the value of array[y] because the value of expression x,y is y.
You likely meant to write

(void)printf("% d ", array[x][y]);

Because the array is a two dimensional array, the value of array[y] is a
pointer to a one-dimensional array. The values being printing in each
row are &array[0][0], &array[1][0], &array[2][0], &array[3][0], and
&array[4][0].

[Irrwahn Grausewitz]

Irrwahn Grausewitz <ir*****@freene t.de> wrote in
<uh************ *************** *****@4ax.com>:

Arrrrrrrrrrrgh, I did it again, stupid me!
<snip>

in turn a pointer to an array of five ints, which you print using

^^^^^^^^^^^^^^^ ^
<delete this>
<snap>

--
Rain is just liquid sunshine.