What is the difference between prefix and postfix operators?

Question

The following code prints a value of 9. Why? Here return(i++) will return a value of 11 and due to --i the value should be 10 itself, can anyone explain how this works?

#include<stdio.h> main() {     int i= fun(10);     printf("%d\n",--i); }  int fun (int i) {     return(i++); } 

Answer 1

There is a big difference between postfix and prefix versions of ++.

In the prefix version (i.e., ++i), the value of i is incremented, and the value of the expression is the new value of i.

In the postfix version (i.e., i++), the value of i is incremented, but the value of the expression is the original value of i.

Let's analyze the following code line by line:

int i = 10;   // (1) int j = ++i;  // (2) int k = i++;  // (3) 

1. i is set to 10 (easy).
2. Two things on this line:
   • i is incremented to 11.
   • The new value of i is copied into j. So j now equals 11.
3. Two things on this line as well:
   • i is incremented to 12.
   • The original value of i (which is 11) is copied into k. So k now equals 11.

So after running the code, i will be 12 but both j and k will be 11.

The same stuff holds for postfix and prefix versions of --.

Answer 2

Prefix:

int a=0;  int b=++a;          // b=1,a=1  

before assignment the value of will be incremented.

Postfix:

int a=0; int b=a++;  // a=1,b=0  

first assign the value of 'a' to 'b' then increment the value of 'a'