Help needed in solving this problem

[sasiraj]

#include<stdio.h>

#define TOTAL_ELEMENTS (sizeof(array) / sizeof(array[0]))
int array[] = {23,34,12,17,204,99,16};

int main()
{
int d;

for(d=-1;d <= (TOTAL_ELEMENTS-2);d++)
printf("%d\n",array[d+1]);

return 0;
}

The above snippet doesnt print the array as expected.
I think the error is in the Test-condition in for loop.
If i cast it to INT [(int)TOTAL_ELEMENTS-2)] than its working fine.
y is it so..can sbd tell me what the return type of sizeof
is??

[Walter Roberson]

In article <1120500850.2...@g44g2000cwa.googlegroups.com>,

sasiraj <vardh...@gmail.com> wrote:
> #define TOTAL_ELEMENTS (sizeof(array) / sizeof(array[0]))

> int d;

> for(d=-1;d <= (TOTAL_ELEMENTS-2);d++)

>I think the error is in the Test-condition in for loop.

>If i cast it to INT [(int)TOTAL_ELEMENTS-2)] than its working fine.
>y is it so..can sbd tell me what the return type of sizeof
>is??

sizeof has type size_t . The exact width of size_t is unspecified,
but one thing that -is- specified about it is that it is unsigned.
You might be having difficulty with comparing a negative int
to an unsigned integral type.
--
"This was a Golden Age, a time of high adventure, rich living and
hard dying... but nobody thought so." -- Alfred Bester, TSMD

[Rouben Rostamian]

In article <1120500850.2...@g44g2000cwa.googlegroups.com>,
sasiraj <vardh...@gmail.com> wrote:

From K&R2, Section A6.5:

[automatic conversion of operands]
If either operand is unsigned int, the other is
converted to unsigned int.

Thus the -1 in d=-1 is converted (unsigned int)(-1) which
generally is a pretty large number.

--
Rouben Rostamian

[ice]

sasiraj wrote:
> #include<stdio.h>
>
> #define TOTAL_ELEMENTS (sizeof(array) / sizeof(array[0]))
> int array[] = {23,34,12,17,204,99,16};
>
> int main()
> {
> int d;
>
> for(d=-1;d <= (TOTAL_ELEMENTS-2);d++)
> printf("%d\n",array[d+1]);
>
> return 0;
> }
>
> The above snippet doesnt print the array as expected.

hello sasi
Do you want to print the array in order as it is declared?If yes they
are printing in order .
and if no then clarify in which order?

[Jaspreet]

sasiraj wrote:
> #include<stdio.h>
>
> #define TOTAL_ELEMENTS (sizeof(array) / sizeof(array[0]))

Just make it:
#define TOTAL_ELEMENTS (signed)(sizeof(array) / sizeof(array[0]))

Try going through sizeof help and you would have the answer yourself.

> int array[] = {23,34,12,17,204,99,16};
>
> int main()
> {
> int d;
>
> for(d=-1;d <= (TOTAL_ELEMENTS-2);d++)

OR another choice could be to initialise d to 0. But I guess you did
not want this. Soo, go through sizeof help.

[Jaspreet]

ice wrote:
> sasiraj wrote:
[snip program]

> > The above snippet doesnt print the array as expected.
> hello sasi
> Do you want to print the array in order as it is declared?If yes they
> are printing in order .
> and if no then clarify in which order?
> > I think the error is in the Test-condition in for loop.
> > If i cast it to INT [(int)TOTAL_ELEMENTS-2)] than its working fine.
> > y is it so..can sbd tell me what the return type of sizeof
> > is??

I dont think its should print the contents without you modifying the
#define preprocessor. Please check once again.

OR probably I need to check my compiler. ;)