pointer giving unexpected value

Question

I am reading Understanding Pointers in c.

I found one program here it is.

#include<Stdio.h>
#include<conio.h>
int main()
{
    static int arr[]={97,98,99,100,101,102,103,104};
    int *ptr=arr+1;
    printf("\nThe base Address is:%u\n",arr);
    print(++ptr,ptr--,ptr,ptr++,++ptr);
    return getch();
}

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

The program is perfect what i think is that it must generate following output: 99 99 98 98 100. But it gives following output: 100 100 100 99 99

I am unable to understand help me to understand it.

Hi @google, please google for your question before asking here. So many duplicates have already been answered here. — Don't You Worry Child, Oct 16 '13 at 11:45
Dan and 0xF1: It is kind of funny (or at least oxymoronic) when the question is asked by a user called google. — Klas Lindbäck, Oct 16 '13 at 13:20

score 4 · Accepted Answer · answered Oct 16 '13 at 11:45

In addition to the order of evaluation of function parameters being unspecified, the program sports several items with undefined behavior:

It tries to access the value of an expression with side effects before reaching the next sequence point (it does that several times in the call of print), and
It tries printing a pointer using %u format specifier.

The way the program is written it can print anything at all, not print anything, or even crash: that's the consequences of having undefined behavior.

If you want the evaluation of parameters to be in a specific order, introduce temporary variables, and do evaluation sequentially, like this:

#include<stdio.h>
#include<conio.h>
int main()
{
    static int arr[]={97,98,99,100,101,102,103,104};
    int *ptr=arr+1;
    printf("\nThe base Address is:%p\n",(void*)arr);
    int *tmp0 = ++ptr;
    int *tmp1 = ptr--;
    int *tmp2 = ptr;
    int *tmp3 = ptr++;
    int *tmp4 = ++ptr;
    print(tmp0, tmp1, tmp2, tmp3, tmp4);
    return getch();
}
print(int *a,int *b,int *c,int *d,int *e)
{
      printf("%d  %d  %d  %d  %d",*a,*b,*c,*d,*e);
}

thanks this is knowledgable answer and i really feeling good to understand this. — Nirav Kamani, Oct 16 '13 at 11:51

score 2 · Answer 2 · edited Oct 16 '13 at 18:48

2

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

is undefined behaviour because the order in which the parameters are evaluated is not specified by the C standard.

edited Oct 16 '13 at 18:48

This isn't my real name

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answered Oct 16 '13 at 11:43

Ingo Leonhardt

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pointer giving unexpected value

2 Answers2