Increment order

Question

I am facing problems in understanding the increment orders in C++.

I am aware that increments are unary operators thus they come after parenthesis from right to left.

My question is when do we increase the number?

Here is a simple code:

#include <iostream>
using namespace std;

int main()
{
  int a1;
  int a(12),b(3);

  a1=7+10%3-5;
  b=a/b++;

  cout<<a1<<"\t"<<b<<endl;
  return 0; 
}

Here I get a=3 that's right but b=5 , I think it is 3 because we start from the right and increase by 1 then 12/4 gives 3.

`b++` increments _after_ getting `b`'s value. `++b` would increment _before_ getting `b`'s value. — Uwe Keim, Mar 21 '19 at 15:45
@UweKeim I think K&R said `b++` increments `b` after the calculation is done, not after getting `b`'s value. — Aykhan Hagverdili, Mar 21 '19 at 15:53
@UweKeim I rechecked the book and it says "But the expression `++n` increments `n` *before* its value is used, while `n++` increments `n` *after* its values has been used." That means I was wrong. Sorry for the confusion — Aykhan Hagverdili, Mar 21 '19 at 16:09
@Bathsheba it doesn't assign anything to itself. It's [K&R](https://en.wikipedia.org/wiki/The_C_Programming_Language) we are talking about here :D The book was just describing what happens when you do `x = n++;` — Aykhan Hagverdili, Mar 21 '19 at 16:32
@Ayxan: Which is well-defined: `x` gets the unincremented `n`. — Bathsheba, Mar 21 '19 at 16:33
hey, thanks for your replies. But what I really wanted to know was when exactly the increment happens. Does it divide 12 by 3 .... get the answer and adds 1 to it? I understood that it adds 1 to b .... divide 12 by 4 and you get 3 !! please correct me if I am missing something. — aysh_99, Mar 21 '19 at 16:33
@Bathsheba what happens when I do `x = 2 * 5 + n++`? Does `n` get incremented immediately after `n++` is evaluated and its value is taken, or does it wait until the end of the calculation (statement)? — Aykhan Hagverdili, Mar 21 '19 at 16:36
`n++` is the original value of `n` for the evaluation of the expression. I would avoid saying *statement* since `x = 2 * 5 + (n++, n)` recovers the pre-increment behaviour because in this guise `n++` *is* the expression. — Bathsheba, Mar 21 '19 at 16:36
@Bathsheba that's what I was looking for. What do you mean by "recovers"? — Aykhan Hagverdili, Mar 21 '19 at 16:39

Bathsheba · Answer 1 · 2019-03-21T16:35:42.937

Note that the C++ grammar implies that the associativity of the postfix increment is from left to right, and that of the prefix increment is right to left.

The behaviour of b = a / b++; is actually undefined. This is because = is not a sequencing point, so there are simultaneous reads and writes on b.

(The same applies to C.)

It's a variant on i = i++;: for more on that see Is the behaviour of i = i++ really undefined?

Increment order

1 Answers1