Newbie Qn

**robertlee** · 12-01-2007

Code:

#include<stdio.h>
int x=21;
int y=25;
main()	{
	clrscr();
	funct(&x);
		}
		funct(int *py)
		{
		++*py++;
		printf("&#37;x %d\n",py,*py);
		getchar();
		}


The output is 196 and 25.
My qn is for :
++*py++; /*This statement means the add py is pointing to increments. but does the value of what py is pointing to get increment too? I was thinking why output isnt 196 and 26 instead.
The output shows that it isnt. please correct me if im wrong about my statement*/

**Elysia** · 12-01-2007

Originally Posted by robertlee

Code:

#include<stdio.h>
/* Don't use global variables */
int x=21;
int y=25;
int main() /* Main should return int, see FAQ */
{ 
	clrscr();
	funct(&x);
	return 0; /* Main should return 0 unless specified otherwise */
}

void funct(int *py) /* All functions must have a return type. If none is required, use void */
{
	/*++*py++;*/ /* Bad! */
	++*py /* Is the same as ++(*py), increments *py by 1 */
	*py++ /* Extremely bad! First dereferences py, then increments the pointer by sizeof(*py) */
	printf("&#37;x %d\n",py,*py);
	getchar();
}

Use proper indentation! This is almost the worst indentation I've seen! I fixed it, so it looks decently.
Stay away from tricky things such as ++*py++. Very bad use. See comments.

**MacGyver** · 12-01-2007

Code:

++*py++;

Is it me or does this appear to have wings on it?

**robertlee** · 12-01-2007

Thanks for your alteration!
actually its not me who wants to use ++*p++;
the source code is from a tutorial from sch actually. and i'm scratching my head on how come *py didnt get incremented.

**Elysia** · 12-01-2007

I can't say, but it's really bad form and should be split up, if possible.

**matsp** · 12-01-2007

When was this tutorial written, 1980?

You should definitely specify return values for your function. You have no prototype for funct, so the compiler has to "guess" what you are up to there. clrscr() is not a portable function [meaning it only works on some systems].

*py is incremented. Before py is incremented. So the value you see when you print *py is the value of "y", whilst the value of x is the one that gets incremented. If you do this:

Code:

#include <stdio.h>

void funct(int *py);

int x=21;
int y=25;
int main()	
{
  funct(&x);
  printf("x = %d y = %d\n", x, y);
  return 0;
  getchar();
}

void funct(int *py)
{
  ++*py++;
  printf("%p %d\n",py,*py);
}

You will see that "x" is 22 instead of 21. I've also changed the %x for printing the pointer to a %x - as modern compilers don't like pointers being printed with %x without warning about converting pointers to integers.

--
Mats

**iMalc** · 12-01-2007

++*py++ is undefined behaviour, meaning it could do anything (set *py to zero, throw an exception, launch notepad, erase your hard disk)

You were also missing the function prototype, matsp added it:

Code:

void funct(int *py);

It shouldn't compile without that.

matsp, you accidentally put getchar after the return

**matsp** · 12-01-2007

Originally Posted by iMalc

++*py++ is undefined behaviour, meaning it could do anything (set *py to zero, throw an exception, launch notepad, erase your hard disk)

You were also missing the function prototype, matsp added it:

Code:

void funct(int *py);

It shouldn't compile without that.

matsp, you accidentally put getchar after the return

That's what you get when you edit the code after you compiled and tested it

Actually, most compilers only give a warning for a missing prototype in C. C++ is slightly different, as it will probably make assumptions with regards to the function signature, and then get a linker error if/when the actual function has a different. It is of course wise to heed the warning in this case, as you can get very "funny" results from mismatched prototype/definitions [or the compilers guess of what the prototype is].

--
Mats

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