#define behaviour

[Bargi]

Hi,

Can any body explain me the output of following:

#include<stdio.h>
int array[]={1,2,3,4,5,6,7,8};
#define SIZE (sizeof(array)/sizeof(int))

int main()
{
printf("%d",SIZE);
if(-1<=SIZE) printf("1");
else printf("2");
return 0;
}

8
2

SIZE is greater than -1 so it should print 2. It's working fine with positive numbers.

Thanks

[Salem]

Do you know what type SIZE is?
Do you know how the compiler resolves comparisons between signed and unsigned types?

[zacs7]

SIZE will always be >= 0 (it's unsigned)

A sizeof expression evaluates to an unsigned value equal to the size in bytes of the "argument" datatype

[Bargi]

Thanks for reply.........
But if it >=0 then why 1 is printed ?
Please explain it in more details ......

Thanks a lot..

[zacs7]

You obviously didn't answer question #2

Do you know how the compiler resolves comparisons between signed and unsigned types?

Remember, when unsigned values go below 0 they loop back to their maximum value, and so on.

Try the following (assuming char has the range -128 to 127, unsigned char 0 to 255):

unsigned char i = 0;
for(i = 0; i < 512; i++)
{
    printf("&#37;d\n, i);
}

You'll probably see it hit 255 and go back to 0 ... 255 then back to 0

Now how do you expect, say -128 to be compared with 255?

[Bargi]

Thanks zacs7...... u solved the problem....

[zacs7]

Actually Salem did, I just answered the question you were supposed to

[shwetha_siddu]

Hi,

Can any body explain me the output of following:

#include<stdio.h>
int array[]={1,2,3,4,5,6,7,8};
#define SIZE (sizeof(array)/sizeof(int))

int main()
{
printf("%d",SIZE);
if(-1<=SIZE) printf("1");
else printf("2");
return 0;
}

SIZE is greater than -1 so it should print 2. It's working fine with positive numbers.

Thanks

Internally your -1 will be stored in 2's complement hence that will be some hex value
hence that will be always greater then SIZE..condition fails and gets else part thats all

[robwhit]

I don't think it's about overflow. There are no operations on numbers except for the like operands of size_t and the relational operator. I think it has to do with integer promotions or something like that. But I'm not really good at that.

[zacs7]

I never said it was about overflow.

[matsp]

Yes, -1 is "the biggest value you can describe" when using it as a unsigned value. So NOTHING will be bigger than -1 in unsigned math.

--
Mats

[robwhit]

zacs7, you were inferring that it was something to do with overflow in post 5, right?

[zacs7]

No, I was mearly outlining the difference between unsigned and signed. ie, giving the user details as they asked. That you can't represent -k to k - 1 and 0 to 2k - 1 with n bits, it's one or the other.

Although I can see how you thought I was talking about overflow