uncertainty in order of execution - any explanation?

**ylzhang** · 05-30-2003

I have a piece of code which prints the values and addresses of a 2D array:

Code:

#include<stdio.h>

int main(void)
{
  /* declare array[3][4] which is stored contigurously in memory */
  int array[3][4]={{1,2,3,4},
	             {5,6,7,8},
                             {9,10,11,12}};

  int *ptr,i,j;
  int nrows=3,ncols=4;

  puts("Using array notation...");
  for(i=0; i<nrows; i++)
    for(j=0; j<ncols; j++)
      printf("\narray[%d][%d]=%d",i,j,array[i][j]);

  puts("\n\n\nUsing pointer notation...");
  /* type cast since ptr is a pointer to int; array is a 
     constant pointer to an int array[4] */
  ptr=(int *)array;
  for(i=0; i<nrows; i++)
    for(j=0; j<ncols; j++)
      printf("\narray[%d][%d]=%d",i,j,*ptr++);

  printf("\n\n");

  ptr=(int *)array;
  for(i=0; i<nrows; i++)
    for(j=0; j<ncols; j++)
      printf("\n&array[%d][%d]=%u",i,j,ptr++);


  printf("\n\n");

  /*  ptr=(int *)array;
  for(i=0; i<nrows; i++)
    for(j=0; j<ncols; j++)
    printf("\narray[%d][%d]=%d at address %u",i,j,*ptr++,ptr++);*/

  ptr=(int *)array;
  for(i=0; i<nrows; i++)
    for(j=0; j<ncols; j++)
      {
	printf("\narray[%d][%d]=%d at address %u",i,j,*ptr,ptr);
	ptr++;
      }

  printf("\n\n");

  return 0;
}

The commented out region did not give me correct result, which
was obtained after I put the pointer increment out of the printf statement:

Not working...

Code:

printf("\narray[%d][%d]=%d at address %u",i,j,*ptr++,ptr++);*/

Working...

Code:

printf("\narray[%d][%d]=%d at address %u",i,j,*ptr,ptr);
ptr++;

However, it was no program when ptr++ or *ptr++ appear singly in a printf statement as in previous examples. I read something about the order of execution of ptr++ and *ptr++ can be uncertain when they are printed together. Any more insightful explanation is much welcome!!! Or any other things similar to this situation that I should look out for?

Thanks in advance!

**Wledge** · 05-30-2003

1. In your commented out region you have 2 (two) post-inc,but only 1 in the other code.
2. You change the value of ptr more than once between two sequence points -> undefined behavior.
3. The format specifer for pointers is p -> undefined behavior.

**WaltP** · 05-30-2003

Code:

printf("\narray[%d][%d]=%d at address %u",i,j,*ptr++,ptr++);*/

When this statement is executed, ptr is incremented after printing the *ptr++ parameter, then incremented again for the ptr++.

Therefore, the second ptr does not have the value expected.

Illiustrated with the following code

Code:

#include<stdio.h>

int xx[] = {11,22,33,44,55,66,77,88,99,38};

int main()
{
    int *p,i;
    
    p = xx;
    for (i=0; i<5; i++)
    {
        // display the index, value, address
        printf("%d)  %d %p \n", i, *p, p);
        p++;
    }
    printf("\n");

    p = xx;
    for (i=0; i<5; i++)
    {
        // display the index, value, address
        printf("%d)  %d %p \n", i, *p++, p++);
    }
    printf("\n");
}

**Dave_Sinkula** · 05-30-2003

Q3.2 - Q3.8 - Q3.3 - Q3.9

**WaltP** · 05-31-2003

Wow! Thanks Dave. I never new there were ideosyncracies with ++ and -- like that. I now wonder if the compilers I've used redefined how the auto-inc worked, or I just never used the ++ in a undefined form!

So bottom line is if a value is inc- or decremented, do not use it elsewhere in the same statement.

**Wledge** · 05-31-2003

Originally posted by WaltP
I never new there were ideosyncracies with ++ and -- like that.

It is not dependent on ++ or --.
This code snippet outputs three different results on three different compilers (two gcc ports).

Code:

int p = 0;
printf("%d %d\n",p += 1,p += 1);

changing a value more than once between two sequence points..

**vikasgp** · 05-31-2003

The order in which arguments to a function are evaluated is not guaranteed. So *never* write code that depends on it.

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