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 ex

Answer 1

fun(10) returns 10. If you want it to return 11 then you need to use ++i as opposed to i++.

int fun(int i)
{
    return ++i;
}

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'

Answer 3

Let's keep this as simple as possible.

let i = 1
console.log('A', i)    // 1
console.log('B', ++i)  // 2
console.log('C', i++)  // 3
console.log('D', i)    // 4

A) Prints the value of i B) First i is incremented then the console.log is run with i as it's new value C) Console.log is run with i at its current value, then i will get incemented D) Prints the value of i

In short if you use the pre-shorthand i.e(++i) i will get updated before the line is executed. If you use the post-shorthand i.e(i++) the current line will run as if i had not been updated yet then i gets increased so ther next time your interpreter comes accross i it will have been increrased.

Answer 4

There are two examples illustrates difference

int a , b , c = 0 ; 
a = ++c ; 
b = c++ ;
printf (" %d %d %d " , a , b , c++);

• Here c has value 0 c increment by 1 then assign value 1 to a so value of a = 1 and value of c = 1

• next statement assiagn value of c = 1 to b then increment c by 1 so value of b = 1 and value of c = 2

• in printf statement we have c++ this mean that orginal value of c which is 2 will printed then increment c by 1 so printf statement will print 1 1 2 and value of c now is 3

you can use http://pythontutor.com/c.html

int a , b , c = 0 ; 
a = ++c ; 
b = c++ ;
printf (" %d %d %d " , a , b , ++c);

• Here in printf statement ++c will increment value of c by 1 first then assign new value 3 to c so printf statement will print 1 1 3

Answer 5

Actually what happens is when you use postfix i.e. i++, the initial value of i is used for returning rather than the incremented one. After this the value of i is increased by 1. And this happens with any statement that uses i++, i.e. first initial value of i is used in the expression and then it is incremented.

And the exact opposite happens in prefix. If you would have returned ++i, then the incremented value i.e. 11 is returned, which is because adding 1 is performed first and then it is returned.

Answer 6

The function returns before i is incremented because you are using a post-fix operator (++). At any rate, the increment of i is not global - only to respective function. If you had used a pre-fix operator, it would be 11 and then decremented to 10.

So you then return i as 10 and decrement it in the printf function, which shows 9 not 10 as you think.