why the printf function showing desired output in the following code?

Question

#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
    int a=1;
    printf("%d\t%d\t%d\n",a,++a,a++);
    return 0;
}

Why the output of the code is 3 3 1. someone explain me how this kind of output happen?

@play store - Did you not get a warning like this : **warning: unsequenced modification and access to 'a' [-Wunsequenced]** ? — Vishaal Shankar, Jun 17 '19 at 08:29

Narase · Answer 1 · 2019-06-17T08:47:52.877

1

Seems like your compiler reads the parameters from right to left

printf("%d\t%d\t%d\n",a,++a,a++); // a = 1

a++ returns a and increments it by 1

printf("%d\t%d\t%d\n",a,++a, 1); // a = 2

++a increments a by 1 and returns the result

printf("%d\t%d\t%d\n",a, 3, 1); // a = 3

a is just a

printf("%d\t%d\t%d\n", 3, 3, 1); // a = 3

But AFAIK this is kinda UB because the c++ standard doesnt rule in which order the parameters are read, so I wouldnt bet on it beeing the same on different compilers

Edit: With C++17 its no longer UB but unspecified. You should still avoid it

edited Jun 17 '19 at 08:47

answered Jun 17 '19 at 08:20

Narase

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please see the comment from [StoryTeller](https://stackoverflow.com/users/817643/storyteller) below my answer. It isn't always UB anymore. In C++17 it is unspecified. – Tarick Welling Jun 17 '19 at 08:32
Ive read it. But "unspecified" is just slightly better than UB. Compiler specific implementations should be avoided too – Narase Jun 17 '19 at 08:46
totally agreed, but your answer becomes better when it is included :) – Tarick Welling Jun 17 '19 at 08:48
your last 2 paragraphs could be reprhased a bit. When there is UB it does not just mean that you get them evaluated in unpredictable order, but UB is something the compiler may assume to never happen, hence before C++17 not only the outcome is undefined but the whole code could do anything, with C++17 you still dont know the order, but there is no UB, which is a fundamental difference – 463035818_is_not_an_ai Jun 17 '19 at 10:14

score 0 · Answer 2 · edited Jun 20 '20 at 09:12

0

This is undefined behaviour as per the order of evaluation. reference (see chapter Undefined behavior)

The output is:

3 3 1

because it evaluates as follows:

a++
use a(1) and a becomes 2
++a
a becomes 3 and use a(3)
use a(3)

Important is to know that a++ is post-increment and ++a is pre-increment. Post-increment means: use the value and increase it after. Pre-increment means: increment the value and use the incremented value.

sidenote: C++17 changed this from undefined behaviour to unspecified but in previous versions it still is undefined behaviour.

edited Jun 20 '20 at 09:12

Community

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answered Jun 17 '19 at 08:19

Tarick Welling

3,119
3
19
44

1

You should read the (since C++17) section more closely. It's no longer UB, just unspecified. – StoryTeller - Unslander Monica Jun 17 '19 at 08:22
the cppreference page is a bit vague on that . (I personally don't see a C++17 section) but indeed it isn't UB anymore. Do you have any updated documentation on the order of evaluation in c++17? – Tarick Welling Jun 17 '19 at 08:26
Easy to miss the mentions of C++17 on mobile. Anyway, the very reference you link to has plenty of examples that say "undefined behavior until C++17". – StoryTeller - Unslander Monica Jun 17 '19 at 08:27
Stupid me, I was expecting a "since C++17" somewhere but of course the statement unspecified is all the documentation needed for that. – Tarick Welling Jun 17 '19 at 08:30

why the printf function showing desired output in the following code?

2 Answers2

sidenote: C++17 changed this from undefined behaviour to unspecified but in previous versions it still is undefined behaviour.