Undefined behavior while using increment operator

Question

I am new to C, i have an Increment operator program in C

#include<stdio.h>
main(){
  int a, b;
  a = 2;
  b = a + ++a + ++a;
  printf("%d", b);
  getchar();
}

The output is 10, can someone explain me how the output will be 10 .

Undefined behavior. Nasal demons. – Fred Larson Jul 21 '11 at 16:44 — Fred Larson, Jul 21 '11 at 16:44
Questions of this sort come up on SO pretty regular. – Andrey Jul 21 '11 at 16:45 — Andrey, Jul 21 '11 at 16:45

score 6 · Answer 1 · answered Jul 21 '11 at 16:44

6

a + ++a + ++a;

Behaviour for this is undefined. The compiler might generated code that evaluated this as 2 + 4 + 4 or 3 + 3 + 4, but any combination/ordering of incrementing and accessing is a "valid" result.

answered Jul 21 '11 at 16:44

Karoly Horvath

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couldn't it also be `4 + 4 + 4` (increments first?) – KevinDTimm Jul 21 '11 at 16:51
1

Not only that, but since the behavior is undefined, printing `42` or `Bad diff! No cookie!` is a valid behavior (just extremely unlikely in practice). So is making the computer burst into flames, or as the classical formulation goes, making demons fly through your nose. – Gilles 'SO- stop being evil' Jul 21 '11 at 16:51
sure, any ordering is fine. I just wrote examples that add up to 10. – Karoly Horvath Jul 21 '11 at 16:52
@Giles the behaviour is *not* undefined, the *order* of the behaviour is undefined. Printing `42` is not allowed. (You were right, however, to correct my use of the term "associative".) – spraff Jul 21 '11 at 16:55
@Gilles: I'm waiting for like 10 years to a compiler that does those things. It looks like I have to write it myself. – Karoly Horvath Jul 21 '11 at 16:55
why the Behaviour for this is undefined? – useCase Jul 21 '11 at 17:15

spraff · Accepted Answer · 2011-07-21T16:53:20.170

3

This is undefined, the ++i can happen in any order.

Function call arguments are also ambigiously evaluated, e.g. foo(++i,++i).

Not all operator chains are undefined, a||b||c is guaranteed to be left-to-right, for example.

The guarantees are made in places known as sequence points although this terminology is being deprecated and clarified in C++0x.

What's odd in your example is that neigher 2+3+4 nor 4+4+3 happened, so the compiler evaluated the left side first in one step and the right side first in the other. This was probably an optimisation to flatten the depencency graph.

edited Jul 21 '11 at 16:53

answered Jul 21 '11 at 16:44

spraff

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It's not just that the `++i` can happen in any order: there's no sequence point around operators (except `&&`, `||`, `?:` and `,`), nor is there one around the `,` that separates function arguments. See e.g. [N1256](http://www.open-std.org/jtc1/sc22/wg14/www/docs/n1256.pdf) §C and §5.1.2.3. If an object is both written and read with no intervening sequence point, the behavior is undefined (§6.5.16.3). – Gilles 'SO- stop being evil' Jul 21 '11 at 17:07

Undefined behavior while using increment operator

2 Answers2