Are all members of following structs and arrays initialized with zero?

Question

Furthermore, is there a difference between the initialization of the variables one and two, and the initialization of the varibles three and four? Background of the Question is, that i get an compiler error in Visual Studio 6.0 with the initialization of variable two and four. With Visual Studio 2008 it compiles well.

struct stTest
{
  int a;
  char b[10];
};

stTest one = {0};
stTest two = {};
stTest three[10] = {0};
stTest four[10] = {};

This all should be valid syntax. What is the nature of the compiler error and can you provide an example that wraps this snippet in a short `main()` that still causes the error? — Adam, Dec 22 '09 at 22:26
Visual Studio 6.0 had serious problems in conforming to the standards. Microsoft has done a lot of work, and VS 2008 is very close to the standard, although there's a few minor issues left. — David Thornley, Dec 22 '09 at 22:27
Visual Studio 6 had no real standards to conform to. It was released before the first C++ standard came out. — AnT stands with Russia, Dec 22 '09 at 22:28

score 2 · Accepted Answer · answered Dec 22 '09 at 22:26

Yes, all of them a required to be initialized with 0 by the language standard (C++98).

Visual Studio 6 is known not to perform the proper handling of {} case: it doesn't even support {} syntax, if I remember correctly.

However, Visual Studio 6 is a pre-standard compiler. It was released before the C++98 standard came out.

score 1 · Answer 2 · edited May 23 '17 at 12:03

1

See Michael Burr's answer to a similar question.

The short answer is yes, but a little emphasis sometimes helps, e.g.,

stTest s = {0};

edited May 23 '17 at 12:03

Community

1
1

answered Dec 22 '09 at 22:42

Greg Bacon

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score 0 · Answer 3 · answered Dec 22 '09 at 22:22

0

The initializations are the same. Visual Studio 2.0 is broken.

Edit:

Yes, they are initialized to zero.

answered Dec 22 '09 at 22:22

Richard Pennington

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score 0 · Answer 4 · answered Dec 22 '09 at 22:38

(Ripped from MSDN)

If I have a structure called tPoint ...

    struct tPoint {   
        int x;   
        int y;
    };

... and I use it as follows ...

tPoint spot {10,20};

... I can expect that members x and y will be initialized.

As for your first question, I'd expect that the array b is not initialized because you only give one value for initialization.

You can initialize the values to zero by default:

struct stTest
{
  int a = 0;
  char b[10] = {0};
};

You can initialize the array like this:

char i[10] = {0};
stTest one = {0, i};

As for why it compiles with VS 2008 and not VS 6.0, VS 2008 probably ignores the empty set and doesn't try to initialize anything.

Are all members of following structs and arrays initialized with zero?

4 Answers4