Very common Expression evaluation

Question

What is the way of working out:

int i=5;
a= ++i + ++i + ++i;

If we go by normal right to left evaluation, the result should be 21 (6 + 7 + 8) .

I remember studying in school, that the answer was 24(8 + 8 + 8)

But I tried it on CodeBlocks, www.ideone.com, ie gcc 4.8.1 compiler and now I get 22. Could someone please explain the reason

Also: http://stackoverflow.com/questions/4176328/undefined-behavior-and-sequence-points — Benjamin Lindley, Aug 05 '13 at 14:10
A nice quote from one of the questions that this is a possible duplicate of: "Use statements which the language standard actually tells you what they will do. Do not use undefined behaviour and then wonder what's going on. " — Ricky Mutschlechner, Aug 05 '13 at 14:12
@RickyMutschlechner Wow! I read your quote and I seemed to remember that it was from me, then I [looked](http://stackoverflow.com/questions/949433/could-anyone-explain-these-undefined-behaviors-i-i-i-i-i-etc#comment757581_949443) and it was, actually :-) That's not bad for being from 2009. — Daniel Daranas, Aug 05 '13 at 14:31

score 0 · Answer 1 · answered Aug 05 '13 at 14:06

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There are no Sequence Points in that expression, so you can't work it out, it is undefined/compiler dependant.

answered Aug 05 '13 at 14:06

Salgar

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score 0 · Answer 2 · edited May 23 '17 at 12:20

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It is as defined by the C/C++ standards an undefined or implementation-defined behavior. The language standards do not specify the behavior, it allows the compiler implementation to choose (and is usually documented).

Also look at another StackOverflow question: Could anyone explain these undefined behaviors (i = i++ + ++i , i = i++, etc…).

What GCC does in your example is the following:

int i = 5;
a = ++i + (++i + ++i);  // it first gets the value of i and increments it two times
a = ++i + (7 + 7);  // it then reads the new value of i, does the addition and save it as a constant in the stack
a = ++i + 14;  // it then gets the value of i and increments it
a = 8 + 14;  // it then reads the new value of i and add it to the constant
a = 22;

edited May 23 '17 at 12:20

Community

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answered Aug 05 '13 at 14:12

Kyle_the_hacker

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"Undefined" is very different from "implementation-defined". In this case, it's undefined. – molbdnilo Aug 05 '13 at 14:15
@molbdnilo: Since the compiler compiles and the program give a results **it is** defined. – Kyle_the_hacker Aug 05 '13 at 14:17
@Kyle_the_hacker: No. One of the consequences of being undefined is that the compiler can do anything, including compile it and give a result. In fact, for undefined behavior, it is almost always the case that it does this, because it is the simplest thing to do (no need to special case it) – Benjamin Lindley Aug 05 '13 at 14:19
It is also very unlikely that a compiler documents what it will do in such a case (and the behavior can be very different with different compiler settings like optimization levels). – Hulk Aug 05 '13 at 14:29

Very common Expression evaluation

2 Answers2