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 lt stdio h gt  main         int i  fun 10       printf   d n  --i      int fun  int i        return i

User · Answer

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.

User · Answer

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

User · Answer

Explanation   Step 1  int fun int   Here we declare the prototype of the function fun     Step 2  int i   fun 10   The variable i is declared as an integer type and the result of the fun 10  will be stored in the variable i   Step 3  int fun int i   return  i       Inside the fun   we are returning a value return i     It returns 10  because i   is the post-increement operator   Step 4  Then the control back to the main function and the value 10 is assigned to variable i   Step 5  printf   d n   --i   Here --i denoted pre-increement  Hence it prints the value 9

User · Answer

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     i is set to 10  easy   Two things on this line    i is incremented to 11  The new value of i is copied into j  So j now equals 11   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 --

User · Answer

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        2 console log  D   i        3  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

User · Answer

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

User · Answer

In fact return  i    will only return 10   The    and -- operators can be placed before or after the variable  with different effects  If they are before  then they will be processed and returned and essentially treated just like  i-1  or  i 1   but if you place the    or -- after the i  then the return is essentailly  return i  i   1    So it will return 10 and never increment it

User · Answer

First  note that the function parameter named i and the variable named i in main   are two different variables  I think that doesn t matter that much to the present discussion  but it s important to know   Second  you use the postincrement operator in fun    That means the result of the expression is the value before i is incremented  the final value 11 of i is simply discarded  and the function returns 10  The variable i back in main  being a different variable  is assigned the value 10  which you then decrement to get 9

User · Answer

It has to do with the way the post-increment operator works  It returns the value of i and then increments the value

User · Answer

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

User · Answer

i   is post increment  The increment takes place after the value is returned

User · Answer

The postfix increment    does not increase the value of its operand until after it has been evaluated  The value of i   is i   The prefix decrement increases the value of its operand before it has been evaluated  The value of --i is i - 1   Prefix increment decrement change the value before the expression is evaluated  Postfix increment decrement change the value after   So  in your case  fun 10  returns 10  and printing --i prints i - 1  which is 9

User · Answer

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

[c] What is the difference between prefix and postfix operators?

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