What place does post increment and post decrement operators have in the rules of operator precedence in c language

Question

As per the information in this link , post increment and decrement operators have first place. And this link says " Take this example:

foo = *p++;

Here p is incremented as a side effect of the expression, but foo takes the value of *(p++) rather than (*p)++, since the unary operators bind right to left ".

But after doing it practically nothing happened as information mentioned in those links.

#include<stdio.h>
#include<stdlib.h>
int main()
{
    int i = 1;
    int *iptr;
    iptr = &i;
    int num = *iptr++;//1) In this expression the value of 'i' is assigned to num. And an attempt of post incrementing the address stored in *iptr is done as side effect.
    printf("Num value: %d\n",num);

    printf("Pointer: %d\n",*iptr);//2) The address of *iptr is not incremented. If it was then the value of 'i' would not be printed instead it would print the incremented address itself.

    printf("Post increment: %d\n",*iptr++);//3) The address of *iptr will be post incremented (which didn't happen in case of num). But for now the value of 'i' will be printed.

    printf("After increment: %d\n",*iptr);//4) Now the incremented address stored in *iptr will be displayed as there is no value assigned to that address.
    return 0;
}

In the above experiment the effect of post increment can be seen only after a statement terminator. But no effect can be seen even after a statement terminator if a post increment is done on the right operand of the assignment operator. EG int num = *iptr++; (as mentioned in the above experiment). So exactly what place does post increment and decrement operators have in the rules of operator precedence.

Answer 1

The problem with your code is that it has undefined behavior: when you point a pointer to a singular local variable, dereferencing incremented pointer yields undefined behavior.

Derefefencing incremented pointers is well-defined for pointers into an array.

int array[] = {1, 2, 3};
int *iptr = &array[0];
int num = *iptr++;

In addition, printing iptr using %d and dereference operator is incorrect: you need to print it using %p , after casting iptr to void* , without dereferencing:

printf("Pointer: %p\n", (void*)iptr);
// No asterisk here -----------^

Now everything works as expected ( demo ).