Non-initialisation of proper array element causes undefined behavior

Question

I made some tests in VSC to check the behaviors of arrays. I´ve encountered in the output of one test the issue, that there was apparently happend undefined behavior despite the array element was defined proper, but just not initialized (with proper i mean the array element was defined with the array itself, not additionally over the bounds of the array which causes well-known undefined behavior).

Here is my code, and the output of it below it:

The issue is about the output of foo[4] which is 8 instead of 0.

#include <stdio.h> 

int main()
{
    int foo[12];
    int i;

    foo[5] = 6; 
    foo[6] = 7;
    foo[7] = 8;
    foo[8] = 9;
    foo[9] = 10;
    foo[10] = 11;
    foo[11] = 12;

    for(i=0 ; i<=11 ; i++)
    {
        printf("foo[%d] = %d\n",i,foo[i]);
    }
}

Output:

foo[0] = 0
foo[1] = 0
foo[2] = 0
foo[3] = 0
foo[4] = 8
foo[5] = 6
foo[6] = 7
foo[7] = 8
foo[8] = 9
foo[9] = 10
foo[10] = 11
foo[11] = 12

Thereafter i tried it else and wanted to see if foo[5] might is influenced also, if i do not initialise it as well, but it wasn´t the case. foo[4] still had the wrong value btw:

#include <stdio.h> 

int main()
{
    int foo[12];
    int i;

    // foo[5] = 6; 
    foo[6] = 7;
    foo[7] = 8;
    foo[8] = 9;
    foo[9] = 10;
    foo[10] = 11;
    foo[11] = 12;

    for(i=0 ; i<=11 ; i++)
    {
        printf("foo[%d] = %d\n",i,foo[i]);
    }
}

Output:

foo[0] = 0
foo[1] = 0
foo[2] = 0
foo[3] = 0
foo[4] = 8
foo[5] = 0
foo[6] = 7
foo[7] = 8
foo[8] = 9
foo[9] = 10
foo[10] = 11
foo[11] = 12

My Question is: Why is happening here undefined behavior at foo[4]? The array is defined proper with 12 elements.

You're not initializing `foo[4]` anywhere in either code snippet. Thus printing `foo[4]` is UB and you're not guaranteed any particular result. — Blaze, Oct 23 '19 at 12:45
@Blaze Yes, i know. But why does it cause undefined behavior either? I´ve defined the array proper. — RobertS supports Monica Cellio, Oct 23 '19 at 12:46
Unless you explicitly initialize local variables (including arrays) their values and contents will be *indeterminate*. In C++ even reading indeterminate values is undefined behavior (unlike C where it only *might* be, so please pick the language you're really programming in). — Some programmer dude, Oct 23 '19 at 12:47
Yes, you've defined an array, but you never defined it's value. That means it has an undefined valued. — NathanOliver, Oct 23 '19 at 12:47
@GuillaumeRacicot i have used gcc.exe, not g++.exe, so i think C. did not tested it with g++.exe btw. — RobertS supports Monica Cellio, Oct 23 '19 at 12:48
@Someprogrammerdude I did not know, that it is causing undefined behavior then. I´ve thought it is just when i go over the bounds of an array. Does the same apply also to normal variables if i do not initialize them? — RobertS supports Monica Cellio, Oct 23 '19 at 13:00
Yes, it's the same for *all* local non-static variables. Global variables or local static variables will be initialized though. — Some programmer dude, Oct 23 '19 at 13:14
@Someprogrammerdude Thanks for the helpful informations. i really appreciate that. — RobertS supports Monica Cellio, Oct 23 '19 at 13:16

score 2 · Answer 1 · edited Oct 23 '19 at 12:54

2

Why is happening here undefined behavior at foo[4]? The array is defined proper with 12 elements

It is because of out of luck in one word.

That is, when you define local array as below.

int foo[12];

each element of foo will be having indeterminate values until explicitly initialized.

edited Oct 23 '19 at 12:54

gsamaras

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answered Oct 23 '19 at 12:48

kiran Biradar

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I did not know this, thank you very much. Do you have any documentation reference to this? Maybe in the Standards? Does the same thing apply to normal variables, when i do not initialise them? – RobertS supports Monica Cellio Oct 23 '19 at 13:06

gsamaras · Accepted Answer · 2019-10-23T14:54:33.933

2

The issue is about the output of foo[4].

Not only.

The issue is with every uninitialized element of your array, which you later access - those elements have indeterminate (garbage) values.

You don't initialize the 5 first elements of your array, but you access them in the for loop, which invokes Undefined Behavior (UB).

You were just (un)lucky that the first four elements of you array happened to be, today, in your machine, at this time, initialized to the value that you would like them to be.

Here is the output I got by running your code online:

foo[0] = 0
foo[1] = 0
foo[2] = 4195741
foo[3] = 0
foo[4] = 0
foo[5] = 6
foo[6] = 7
foo[7] = 8
foo[8] = 9
foo[9] = 10
foo[10] = 11
foo[11] = 12

edited Oct 23 '19 at 14:54

answered Oct 23 '19 at 12:49

gsamaras

71,951
46
188
305

Thank you very much for this information. I did not know that this is causing undefined behavior. Does the same thing apply to not initialised (normal) variables? – RobertS supports Monica Cellio Oct 23 '19 at 13:03
There I was a time I wasn't aware of that either @RobertS, and that's why I upvoted your question! :) As for the local (the normal as you say) variables, the same philosophy applies too! Check this [What happens to a declared, uninitialized variable in C? Does it have a value?](https://stackoverflow.com/questions/1597405/what-happens-to-a-declared-uninitialized-variable-in-c-does-it-have-a-value). There, you'll find references to the Standard too. Hope that helps! – gsamaras Oct 23 '19 at 14:56
In C89, an indeterminate value was defined as either an unspecified value or a trap representation, and the notion that accessing indeterminate values invoked UB was confined to a non-normative annex describing *general* constructs that may invoke UB. Without knowing whether `int` has trap representations, one couldn't know whether the above would have defined behavior. On platforms where `int` has no trap representations, however, the above would output Unspecified values. – supercat Oct 23 '19 at 15:19

Non-initialisation of proper array element causes undefined behavior

2 Answers2