Strange output ---need help

[jack_2205]

#include <stdio.h>

main()
{
int arr[]={ 0,1,2, 3, 4,5, 6, 7, 8, 9};
int *ptr=arr+1;
print( ++ptr,ptr--,ptr,ptr++,++ptr);
}

print(int *a, int *b, int *c, int *d, int *e)
{
printf("\n%d , %d , %d , %d , %d \n", *a,*b,*c,*d,*e );
}

The output is as follows
2 , 2 , 3 , 1 , 3

Can anyone help me to understand the output.

Thanks..Jack

[drharv]

What is this line supposed to do?

>>> print( ++ptr,ptr--,ptr,ptr++,++ptr);

If you want to print, use printf("%d, %d, %d, %d, %d", ++ptr, ptr--, ptr, ptr++, ++ptr);

Also, this line makes no sense, it's like you are trying to declare pointers in a print statement.

>>> print(int *a, int *b, int *c, int *d, int *e)

The only line giving output is

>>> printf("\n%d , %d , %d , %d , %d \n", *a,*b,*c,*d,*e );

which just happens to grab info from random memory addresses since you never initialized the pointers.

This is just what I think, I could be wrong.

[dbaryl]

First of all, I'm guessing that this is not something that you wrote, but rather a problem you have to figure out (I just don't see what you would be trying to do this way =)

As to what this chunk of code does:

#include <stdio.h>

void print(int *a, int *b, int *c, int *d, int *e);

int main()
{
   // creates an array of integers
   int arr[] = {0, 1, 2, 3, 4,5, 6, 7, 8, 9};

   // ptr gets the address of the second item in the array 
   int *ptr = arr + 1;

   // this sends the addresses of items before and after ptr
   print(++ptr, ptr--, ptr, ptr++, ++ptr);

   return (0);
}

void print(int *a, int *b, int *c, int *d, int *e)
{
   printf("\n%d , %d , %d , %d , %d \n", *a, *b, *c, *d, *e);
}

Now, as to what the code does. You have declared an array of ints, [0-9], and then assigned the address of the second element to ptr. The "++" and "--" either add or subtract one from that value, and based on the position of them, this is done before or after the value is passed down to the function.

When the "++" or "--" is placed BEFORE the variable, 1 is added/subtracted from the variable BEFORE the expression is evaluated. In the case where it is placed AFTER the variable, 1 is added/subtracted from the variable AFTER the expression is evaluated. Try an explanation under link below:
http://www.nmr.mgh.harvard.edu/C/incr_decr.shtml

If seems to me that some of this might be compiler-specific because I got different output on M$VC++ from that of DevC++, and different from yours.

>>which just happens to grab info from random memory addresses since you never initialized the pointers.
I'm pretty sure the pointers are initilized in the line:"print(int *a, int *b, int *c, int *d, int *e)"

[dbaryl]

One more thing, can someone run the code through their compiler and tell me what you get? 'Cause I got 2 different outputs from M$VC and my other compiler... don't know what's wrong...

[hk_mp5kpdw]

I've got MSVC++ 6.0 and I got:
3, 2, 3, 2, 3

[Shiro]

I've got GCC and my result:

3, 3, 3, 2, 2

Edit:

Replace the print in main with:

print(++ptr, ++ptr, ++ptr, ++ptr, ptr);

And the result will be:

5, 4, 3, 2, 1

The reason for this is that variable passing starts with the first last parameter, in this case it is ptr. Then ptr is increased each time.

Assume ptr has address 0000. The print can be seen as:

print (0016, 0012, 0008, 0004, 0000);

Since 0000 is the address of (arr+1), this means that *(0000) = 1.
And that's why you get 5, 4, 3, 2, 1.

Back to the original code:

print(++ptr, ptr--, ptr, ptr++, ++ptr);

First 2 is passed. Since this is *((arr+1)+1). Then again 2 is passed, since it is ptr++ and not ++ptr. After 2 is passed, the pointer is increased. So now *((an+1)+2) == 3 is passed. Then again 3 is passed and ptr is decreased. Then ptr is increased so it ends at the same as where it was, which is 3.

[Salem]

> print( ++ptr,ptr--,ptr,ptr++,++ptr);
Such garbage as this has never had a meaningful outcome, so it really doesnt matter what codemangler 7.2 makes of it, its still broken code

http://www.eskimo.com/~scs/C-faq/q3.1.html

[dbaryl]

Thanks everyone, that clears it up for me. I had no idea that the last parameter is passed first... I thought that this kind of code was just pointless, but the reason I looked at it is because I thought that was an excercise, something to understand rather than use... Also, I guess different compilers interpret it differently...

Well, thanks again.