5

Some of this may be a duplicate, but I am sorry for that.
Let's say I have this struct:

struct foo
{
    int a; 
    int b; 
    int c;
};

1. If struct foo type object is declared in the way that it has automatic storage duration and without initializers, is it guaranteed that all it's members will be force initialized to zero?

{
    // other stuff
    struct foo bar;
    // other stuff
}

2. If struct foo type object is declared in the way that it has automatic storage duration and with some initializers, is it guaranteed that members, which are not explicitly initialized, will be force initialized to zero?

{
    // other stuff
    struct foo bar = {.a = 1}; 
    // other stuff
}

3. If struct foo type object is declared in the way that it has automatic storage duration and by using compound literal expression, is it guaranteed that members, which are not explicitly initialized, will be force initialized to zero?

{
    // other stuff
    func((struct foo){.a = 1});
    // other stuff
}

Any C standard references are much appreciated! Thank you!

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2 Answers 2

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6

Summary, TL;DR:

Storage duration explained:

  • A variable declared inside a function has automatic storage duration (including parameters to functions).
  • A variable declared as static, or a variable declared outside functions at file scope ("global") has static storage duration.

Struct (and array) initialization explained:

  • If you initialize no member and the struct has automatic storage duration, nothing is initialized.
  • If you initialize no member and the struct has static storage duration, all members are zero-initialized.
  • If you initialize any member(s), those you didn't touch get initialized to zero.

The relevant part of the C standard (C17 6.7.9 §10):

If an object that has automatic storage duration is not initialized explicitly, its value is indeterminate. If an object that has static or thread storage duration is not initialized explicitly, then:

  • if it has pointer type, it is initialized to a null pointer;
  • if it has arithmetic type, it is initialized to (positive or unsigned) zero;
  • if it is an aggregate, every member is initialized (recursively) according to these rules, and any padding is initialized to zero bits;

Where "artihmetic type" is standard gibberish for plain variables like int, and "aggregate" is standard gibberish for arrays and structs.

Further down in the same chapter (C17 6.7.9 §19):

...all subobjects that are not initialized explicitly shall be initialized implicitly the same as objects that have static storage duration.

Answers to your questions:

  1. If struct foo type object is declared in the way that it has automatic storage duration and without initializers, is it guaranteed that all it's members will be force initialized to zero?

No, it is not guaranteed; their values are indeterminate as stated in the first sentence of the quote above.

  1. If struct foo type object is declared in the way that it has automatic storage duration and with some initializers, is it guaranteed that members, which are not explicitly initialized, will be force initialized to zero?

Yes, as per C17 6.7.9 §19 cited above.

  1. If struct foo type object is declared in the way that it has automatic storage duration and by using compound literal expression, is it guaranteed that members, which are not explicitly initialized, will be force initialized to zero?

Yes, since compound literals are arrays or structs, they follow the same initialization rules.

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Comments

3

First of all, uninitialized variables with automatic storage will never be initialized. From The C11 standard (ISO/IEC 9899:2011 §6.7.9/10):

If an object that has automatic storage duration is not initialized explicitly, its value is indeterminate.

Then from this structure initialization reference:

All members that are not initialized explicitly are initialized implicitly the same way as objects that have static storage duration.

And if we follow the "initialized implicitly" link we have:

objects with static and thread-local storage duration are initialized as follows

  • ...
  • objects of integral types are initialized to unsigned zero
  • ...

So to answer your questions:

  1. No, no initialization is done as there is no explicit initialization in your code (see the quote from the standard)
  2. Yes, since this is a structure initialization
  3. Yes, for same reason as 2.

The links provided have references to the standard.

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8 Comments

@ryyker Not really, because 1 isn't a structure initialization. For an initialization you need to actually initialize the variable.
I have checked the same place in C reference, but that does not say anything about storage duration of the main declaration of the struct. That got me not sure. "All members that are not initialized explicitly are initialized implicitly the same way as objects that have static storage duration." Does that mean this applies to both statics and autos?
@ryyker As soon as you initialize any object, those objects you don't touch get implicitly initialized. But if you initialize no object (of an automatic storage struct), none are initialized.
@KarolisMilieška No, any uninitialized object with automatic storage duration stays uninitialized. Added quote from the standard. For number 2 and 3 there is an explicit (structure) initialization which changes the semantics.
@Someprogrammerdude That last comment isn't correct. An uninitialized object with static storage duration is always initialized no matter what the programmer does.
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