What is the long version of ={0} in C?

Question

I have been asked this question:

= { 0 } is a "shortcut" for what longer code?

So after many searches and rummaging through my C book I know everything that it does with aggregate initialization but what if I wanted to write the longer version of this code. Here is the actual code from the program just in case

static char vid[25][100] = { 0 } ;

The example you're showing is redundant. static linkage as shown defaults to zero-initialized anyway, so it is somewhat pointless. — WhozCraig, Apr 19 '13 at 03:55
@MichaelBurr LOL!Indeed,one can make it as bloated as one wants. — Rüppell's Vulture, Apr 19 '13 at 04:06

WhozCraig · Answer 1 · 2013-04-19T06:23:50.593

The long version (and you'll be hard pressed to get longer):

static const char vid[25][100] =
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};

That said, the decl is static and therefore is placed in zero-initialized memory to begin with, so this is all somewhat redundant, but at least you know what it looks like.

Replacing all `0` with `'\0'` would make it considerably longer... But nice effort! — Floris, Apr 19 '13 at 05:40
+1 for the effort, but wait did anybody check if this is correct :) — Jens Gustedt, Apr 19 '13 at 06:04
another idea to make this longer, use designated initializers: `{ [0] = { [0] = 0, [1] = 0, [2] = 0 ... } ... }` — Jens Gustedt, Apr 19 '13 at 06:08
@JensGustedt Its correct. 25 rows of 100 elements each =P, and I love the idea of index initializers. Wasn't that a feature add in C99? — WhozCraig, Apr 19 '13 at 06:18
@WhozCraig, yes this was added in C99 and remained in C11, so it is valid C. — Jens Gustedt, Apr 19 '13 at 07:49
After thinking about this some more I agree with you that this is actually the answer that best reflects what `{0}` is shorthand for. — Floris, Apr 19 '13 at 12:18
Well I suppose that writing out the array with NULL would be the long version. — IModulo5, Apr 20 '13 at 03:58

score 6 · Answer 2 · answered Apr 19 '13 at 03:52

6

The standard does not define how the zero'ing is done so it's implementation specific. In practice, for local variables compilers generally generate a memset/RtlZeroMemory/etc. for this. For static/globals no code is required because the variable will be placed into zero initialised memory.

answered Apr 19 '13 at 03:52

Mike Kwan

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I quoted from the same article in my comment above. Of course code is needed to zero the memory - the instructions are executed by the loader, when it sets aside (based on instructions in the executable) the .bss block and ensures that its content is zero. Clock cycles are used. Instructions are carried out. By my definition that is code. Just not code the end user had to write. The memory doesn't "just know". IT IS TOLD. – Floris Apr 19 '13 at 05:24
I see your argument now. If I wanted to be pedantic and argue for the sake of it I might say that .bss is often implemented with copy-on-write pages so the memory might never even be allocated unless touched. – Mike Kwan Apr 19 '13 at 05:27
Glad we are on approximately the same page now. The COW argument is interesting a bit far fetched - especially for a 250 element array. As you recognized in your comment. – Floris Apr 19 '13 at 05:38

score 3 · Answer 3 · answered Apr 19 '13 at 03:53

3

It initializes all the elements of the array to 0 in one go.Else you can imagine the strain doing it individually for all of them.

answered Apr 19 '13 at 03:53

Rüppell's Vulture

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"all in one go"? How clever. I thought memory was accessed sequentially. – Floris Apr 19 '13 at 04:28
@EdS. What I meant is the user(Chad) can initialize all the elements to 0 without having to write a lot of code for that purpose,ie, "in one go".I was not explaining to him how it is done internally.In fact,the user(Chad) didn't even ask for how it is done internally.If he is a beginner,statements like O(1) would only freak him more over a question this simple. – Rüppell's Vulture Apr 19 '13 at 04:40
@Floris I used "In one go" in the context of the user.That is, all he has to do is use ={0} and bingo, all elements are initialized to 0 "at once".I didn't mean all memory is accessed at once just like that.Hope you got what I am trying to say. – Rüppell's Vulture Apr 19 '13 at 04:43
I get it. Just using the `static` keyword is actually all the code needed. But if you have a large block, it will take many clock cycles to be set to zero - that was the only point I was trying to make. – Floris Apr 19 '13 at 05:27
@Floris Yeah indeed, static storage class initializes to 0 by default.But let me see what you and Mike Kwan were debating about. – Rüppell's Vulture Apr 19 '13 at 05:39

Cody Gray - on strike · Accepted Answer · 2013-04-19T04:01:35.200

The long version is a loop that initializes all of the members of the array:

for (int i = 0; i < ARRAY_SIZE; ++i)
{
    myArray[i] = 0;
}

The compiler will probably generate a call to the memset function, which may or may not contain a more optimized version of this naive loop.

But like Dietrich points out, in the example you've shown in the question, you're declaring a static array. That makes things a bit more complicated. Uninitialized static data is placed in a special section of the executable called the .bss section and is required by both the C and C++ standards to be default-initialized to zero. If you explicitly initialize the data, the linker puts it in the .data section of the executable. Explicitly initializing to 0 is a bit of a special case. Smart linkers will omit the explicit initializer and place the data in the .bss section. Dumb linkers may still place the data in the .data section. So it's probably better not to explicitly zero-initialize objects with static linkage.

It will probably not generate `memset()` because the variable is static. — Dietrich Epp, Apr 19 '13 at 03:54
You say that the long version (which the question is about) would be your code, which is using assignment. The correct answer is as given by WhozCraig, explicit initialization of all the members. — Jens Gustedt, Apr 19 '13 at 06:14

score -1 · Answer 5 · edited Jun 20 '20 at 09:12

-1

"Approximayely" the "really long" version of what you are doing is:

static char vid[25][100];
static firstTime = 1;

if(firstTime) {
  int ii, jj;

  for(ii = 0; i < 25; ii++) {
    for (jj = 0; jj < 100; jj++) {
      vid[ii][jj] = 0;
    }
  }
  firstTime = 0;
}

This takes account of the fact that this should only be done the very first time, since the variable is declared as static. In reality, as pointed out in the comments, the static keyword implies initialization to zero, and mapping into a special area called the .bss segment. This gets set to zero by the program loader, or some other mechanism, before main() is even called. Thus the code that is run doesn't look much like the above (zeroing blocks of memory can be done quite efficiently on most processors and doesn't involve looping over multiple dimensions!).

But:

Initialization is something that happens before the main function is called, while my "approximately equivalent" code actually is run during execution; so, as Jens pointed out, it is not really the same thing. WhozCraig actually shows the explicit initiozation for which {0} is shorthand; it is the best answer.

edited Jun 20 '20 at 09:12

Community

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answered Apr 19 '13 at 03:54

Floris

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This is simply wrong. Static means it is zero initialised already and there is no runtime cost. – Mike Kwan Apr 19 '13 at 03:59
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@Mike - OP was asking for an explanation of what is actually happening. I thought this explained that the compiler sets the memory locations used by the static array to zero the first time the code is called. The zeroing happens... even if you don't include the `= {0}`. But if you think about it - when the compiler decides to set aside a chunk of memory for the static array, it DOES have to loop through and set elements to zero. Otherwise there is no way of making sure that it is, in fact, zero. The command that is used may not look much like the above loop, and you never see it... – Floris Apr 19 '13 at 04:03
From Wikipedia entry on .bss: "Typically, the program loader initializes the memory allocated for the bss section when it loads the program. Operating systems may use a technique called zero-fill-on-demand to efficiently implement the bss segment (McKusick & Karels 1986). In embedded software, the bss segment is mapped into memory that is initialized to zero by the C run-time system before main() is entered." – Floris Apr 19 '13 at 04:07
@MikeKwan: How does initialization work? Is it just magic? The memory just appears, instantaneously, set to zero, with no need to check? – Benjamin Lindley Apr 19 '13 at 04:08
@BenjaminLindley - Mike is right to say "no **runtime** cost" as it is the program _loader_ that makes sure the memory is set to zero. But that is a technicality. The processor does execute statements that set memory to zero after you say "I want to run this program" and before the static variable is referenced for the first time. – Floris Apr 19 '13 at 04:14
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@Floris: No, he is not right. The time spent loading your program is part of your program's runtime. If your program takes 20 seconds to load, but the actual main function only takes 1 second, and my program takes 1 second to load, but then does the same job in 3 seconds, nobody would argue that your program is faster than mine. – Benjamin Lindley Apr 19 '13 at 04:20
@BenjaminLindley - fair point. I tried to make the distinction that "run time" starts when `main` is called, but your definition makes more sense. On the whole, I think you and I agree about what actually happens. – Floris Apr 19 '13 at 04:25
@Jens - I understand your point. The _effect_ of my code (setting all elements of the array to zero at the start of the program) is the same, but I agree that WhozCraig nailed it (and I loved your indexed initializer construct - I didn't know that was valid syntax but will remember!) – Floris Apr 19 '13 at 12:09

What is the long version of ={0} in C?

5 Answers5

But:

Linked