Precedence of postfix ++ and prefix ++ in C/C++

Last Updated : 03 Oct, 2022

In C/C++, precedence of Prefix ++ (or Prefix –) has same priority than dereference (*) operator, and precedence of Postfix ++ (or Postfix –) is higher than both Prefix ++ and *. If p is a pointer then *p++ is equivalent to *(p++) and ++*p is equivalent to ++(*p) (both Prefix ++ and * are right associative).

Program 1:

For example, program 1 prints ‘h’ and program 2 prints ‘e’.

C++

// C++ program to explain the precedence of 'Prefix ++'
 
#include <iostream>
using namespace std;
 
int main()
{
    char arr[] = "geeksforgeeks";
    char* p = arr;
    ++*p;
    cout << *p;
    return 0;
}
 
// This code is contributed by sarajadhav12052009

C

// Program 1
#include<stdio.h>
int main()
{
  char arr[] = "geeksforgeeks";
  char *p = arr;
  ++*p;
  printf(" %c", *p);
  getchar();
  return 0;
}

Output

Program 2:

C++

// C++ Program to explain the precedence of 'Postfix ++'
 
#include <iostream>
 
using namespace std;
 
int main()
{
    char arr[] = "geeksforgeeks";
    char* p = arr;
    *p++;
    cout << *p;
 
    return 0;
}
 
// This code is contributed by sarajadhav12052009

C

// Program 2
#include<stdio.h>
int main()
{
  char arr[] = "geeksforgeeks";
  char *p = arr;
  *p++;
  printf(" %c", *p);
  getchar();
  return 0;
}

Output

Program to tell a person’s surname

C++

// C++ program that tells a person's last name
 
#include <iostream>
using namespace std;
 
int main()
{
  string fullName[] = {"Joe", "Donaldson"};
  string *ptr = fullName;
   
  cout << *ptr << "'s Last Name is ";
  *ptr++;
  cout << *ptr << endl;
}
 
// This code is contributed by sarajadhav12052009

Output

Joe's Last Name is Donaldson

Here are some of the programs along with the concept which will help you in understanding the concept better.

When pre/post increment/decrement are used in statements with more than one operand then you have to take care of some rules.

Rule 1. When evaluating pre-increment value should be incremented first and will be assigned at the end of evaluating all the operands.

example :

C++

#include <iostream>
using namespace std;
 
int main() {
 
    int a=1;
      cout<< ++a + a++ ;
}

Now take some time and think of the output of the above program.

If You are thinking that the output should be 4 then you are wrong. The output will be 5.

Here’s the explanation

1. pre-increment increments the value first and then assigns it. here ++a is incremented to 2 but is not assigned because a++ is remaining.

2. Then a++ is assigned 2 and then it is incremented to 3. Now the value of a is 3.

3. This final value will be assigned to ++a because pre-increment values are assigned at the last.

4. So the output will be 3+2= 5.

If you think that you understood the concept then try the next example:

What do you think can be the answer to this example?

C++

#include <iostream>
using namespace std;
 
int main() {
 
    int a=1;
      cout<<a++ + ++a;
}

If you think what is the difference between this and the previous example then I must tell you that the difference is in the position of pre-increment and post-increment. And the output will also be changed. You can run this program to check the output.

The output will be 4.

Here’s the explanation:

1. First the post-increment will be evaluated. It will be assigned 1 and then incremented to 2.

2. then pre- increment will be incremented to 3 and then assigned to 3.

So the sum of the assigned values is 1+3 = 4