int x; int y; int *ptr; is not initialization, right?

Question

I'm reading 'C++ All-in-One for Dummies' by J. P. Mueller and J. Cogswell and stumbled onto this:

#include <iostream>
using namespace std;
int main()
{
    int ExpensiveComputer;
    int CheapComputer;
    int *ptrToComp;
...

This code starts out by initializing all the goodies involved — two integers and a pointer to an integer.

Just to confirm, this is a mistake and should read '... by declaring', right? It's just strange to me that such basic mistakes still make their way to books.

Yes you declare those variables, and define them. What you *don't* do is initialize them. — Some programmer dude, Nov 20 '18 at 11:11
You are correct, and yes, unfortunately there are a lot of crappy books out there. — 500 - Internal Server Error, Nov 20 '18 at 11:11
@molbdnilo, a clear and easy-to-understand explanation that's factually correct, though. :) Writing 'for dummies' books does not justify factual mistakes. — John Allison, Nov 20 '18 at 11:31
Re: should read '... by declaring'" -- that's **defining**. These three statements create three variables. They are definitions. A definition is also a declaration, but a declaration is not a definition. — Pete Becker, Nov 20 '18 at 15:35
Also see [Has C++ standard changed with respect to the use of indeterminate values and undefined behavior in C++14?](https://stackoverflow.com/q/23415661/1708801) — Shafik Yaghmour, Nov 21 '18 at 02:41
Related: https://stackoverflow.com/questions/6032638/default-variable-value — MyStackRunnethOver, Dec 04 '18 at 21:39

score 23 · Accepted Answer · edited Jun 20 '20 at 09:12

23

From the point of view of the language, this is default initialization. The problem is, they are initialized to indeterminate values.

otherwise, nothing is done: the objects with automatic storage duration (and their subobjects) are initialized to indeterminate values.

Default initialization of non-class variables with automatic and dynamic storage duration produces objects with indeterminate values (static and thread-local objects get zero initialized)

Note that any attempt to read these indeterminate values leads to UB.

From the standard, [dcl.init]/7

To default-initialize an object of type T means:

If T is a (possibly cv-qualified) class type ([class]), constructors are considered. The applicable constructors are enumerated ([over.match.ctor]), and the best one for the initializer () is chosen through overload resolution ([over.match]). The constructor thus selected is called, with an empty argument list, to initialize the object.

If T is an array type, each element is default-initialized.

Otherwise, no initialization is performed.

edited Jun 20 '20 at 09:12

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answered Nov 20 '18 at 11:21

songyuanyao

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There is a crucial difference between _initialized with an indeterminate value_ and _not initializated_: the former requires a memory store, the latter does not. – Maxim Egorushkin Nov 20 '18 at 11:34
@MaximEgorushkin That means the expression of cppreference.com is not accurate either..? – songyuanyao Nov 20 '18 at 11:39
Not sure which expression you refer to. – Maxim Egorushkin Nov 20 '18 at 11:41
@MaximEgorushkin *the objects with automatic storage duration (and their subobjects) are initialized to indeterminate values.* – songyuanyao Nov 20 '18 at 11:42
The expression is accurate. There is a difference between _initialized to_ and _initialized with_ (the one you used). – Maxim Egorushkin Nov 20 '18 at 11:43
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@MaximEgorushkin -- that may be formally true, but you cannot write a conforming program that can tell whether a memory store occurred in these cases, so the as-if rule says that they're the same thing. – Pete Becker Nov 20 '18 at 15:33
@PeteBecker: Indeed we can go further: The standard says "initialized to an indeterminate value" specifically to allow compilers to skip the memory store (as opposed to zero-initialization, which would not so allow). You can't have "not initialized" at the level of the C abstract machine (unless you are trying to sneak UB in via the "anything not defined is undefined" rule), so you have to say "initialized with indeterminate values" instead. – Kevin Nov 20 '18 at 18:16
It's actually worse. They are not initialized, and people have posted examples on stackoverflow where the variables took a value that was outside the range of possible values for a variable of that size. – Joshua Nov 20 '18 at 20:30

gsamaras · Answer 2 · 2018-11-20T13:18:57.407

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Yes you are correct.

You declared and defined these variables, you did not initialize them!

PS: What is the difference between a definition and a declaration?

edited Nov 20 '18 at 13:18

answered Nov 20 '18 at 11:13

gsamaras

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https://stackoverflow.com/questions/1410563/what-is-the-difference-between-a-definition-and-a-declaration *defines* – Fantastic Mr Fox Nov 20 '18 at 11:19
@FantasticMrFox, yes, I read that before posting my question. It was just weird to me that the authors, having such experience (for what it's worth), could write something that's so trivially wrong. Having read **songyuanyao** 's answer, it's not so trivial, though. – John Allison Nov 20 '18 at 11:25
FantasticMrFox you are right, updated. @JohnAllison I am not sure, it they wanted to be so on point, they would say default initialize with garbage/random values. Saying initialize a variable and not doing so is **so misleading** and **wrong**. I see what extra information songyuanyao's answer brought to the table, but it's not a boolean one. Anyway, glad you found an answer that fits your thoughts. – gsamaras Nov 20 '18 at 14:29

Maxim Egorushkin · Answer 3 · 2018-11-20T11:32:55.287

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This code both declares and defines three variables but does not initialize them (their values are said to be indeterminate).

A variable declaration only must include keyword extern.

edited Nov 20 '18 at 11:32

answered Nov 20 '18 at 11:18

Maxim Egorushkin

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Lightness Races in Orbit · Answer 4 · 2018-11-20T14:15:55.023

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Right. Hence, "dummies". :)

We can't even blame this on legacy; historically C programmers would declare* a variable and then "initialize" it later with its first assignment.

But it was never the case that simply declaring a variable, without an initializer, were deemed to be "initializing" it.**

So the wording is just wrong.

_{* Technically we're talking about definitions, but when we say "declare a variable" we almost always mean defining declarations.}

_{** Though objects with static storage duration do undergo their own zero-initialisation phase before anything else happens, so forgoing initialisation yourself is not a catastrophe in that case. Still, we cannot claim that we have initialised that object.}

edited Nov 20 '18 at 14:15

answered Nov 20 '18 at 11:46

Lightness Races in Orbit

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A variable _declaration_ only must include keyword `extern`, otherwise it is a _definition_. The initializer is a separate concern. – Maxim Egorushkin Nov 20 '18 at 11:47
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@MaximEgorushkin Definitions are declarations too. But, sure, we can say "simply defining a thing" too and the lesson stays the same. Not really the key point, is it? – Lightness Races in Orbit Nov 20 '18 at 11:53
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In a declaration only you can write, for example, `extern int a[];` (no array bounds), unlike in a definition. Confusing the two never helps. – Maxim Egorushkin Nov 20 '18 at 12:10
@MaximEgorushkin Again, a definition _is_ a declaration. Your own answer says this. – Lightness Races in Orbit Nov 20 '18 at 12:22
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*But it was never the case that simply declaring a thing without an initializer were deemed "initializing" it.* - Except when it is static or global. – Fantastic Mr Fox Nov 20 '18 at 14:01
@FantasticMrFox True :D Well, ish. You're still not initialising those, though thanks to the early zero-init pass you could sort of pretend ;) – Lightness Races in Orbit Nov 20 '18 at 14:09
I've made a few changes to address all your concerns, though the basic point stands. – Lightness Races in Orbit Nov 20 '18 at 14:17
@LightnessRacesinOrbit As a side note, why do you use a picture of a famous person as your profile picture? – Maxim Egorushkin Nov 20 '18 at 14:32
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@MaximEgorushkin Way off-topic and why not? Why is it important to you? Why does gsamaras use a picture of an airplane as their profile picture? – Lightness Races in Orbit Nov 20 '18 at 14:35
@MaximEgorushkin My profile picture is of a famous person also! wooohoooo – Fantastic Mr Fox Nov 20 '18 at 14:40
@FantasticMrFox It's fantastic – Lightness Races in Orbit Nov 20 '18 at 14:41
Certainly there is often some looseness with wording in informal context, but the formal terminology around this is the same in C as in C++. An assignment that is not part of the declaration of a variable is not an "initialization" in C. Informally, of course, the first assignment to a variable can be referred to as an initialization much more broadly than just in C (and C++!) context. – John Bollinger Nov 21 '18 at 15:44

int x; int y; int *ptr; is not initialization, right?

4 Answers4