Initializing a two dimensional array with class static const integers

Question

I'm declaring a three-dimensional array as a class member, using static const class members as the first two bounds:

class A
{
    static const uint8_t screenWidth = 256;
    static const uint8_t screenHeight = 240;

    uint8_t buffer[screenHeight][screenWidth ][3];
}

in Visual Studio 2019 I get the following (weird?) error:

Error (active)  E0098   an array may not have elements of this type

if I resort to the "enum hack" in order to declare class-local compile time integer constants it works:

class A
{
    enum { SH = 240, SW = 256};
    uint8_t buffer[SH][SW][3];
}

shouldn't the former example be C++11 compliant code? (I guess Visual Studio 2019 compiler is).

@stark `const` and `constexpr` have the same effect in this case. Integral types have special rules. — walnut, Mar 05 '20 at 14:38

score 4 · Answer 1 · answered Mar 05 '20 at 12:05

4

I think that an object of the type uint8_t is unable to contain the value 256.:)

Why not just to use the type size_t instead of the type uint8_t?

static const size_t screenWidth = 256;
static const size_t screenHeight = 240;

answered Mar 05 '20 at 12:05

Vlad from Moscow

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@Luca In general it is a bad idea. From the C++ Standard "If the initializer-clause is an expression and a narrowing conversion (8.5.4) is required to convert the expression, the program is ill-formed" In your case the value is truncated to 0. – Vlad from Moscow Mar 05 '20 at 12:16

score 2 · Accepted Answer · answered Mar 05 '20 at 12:07

2

The problem you have is that, in the declaration:

static const uint8_t screenWidth = 256;

the value 256 is not valid for a uint8_t type (range is 0 thru 255), and it 'rolls over' to give an actual value of zero - which is invalid for an array dimension.

Make your dimension 'constants' bigger types, and your code will work:

class A {
    static const uint16_t screenWidth = 256;
    static const uint16_t screenHeight = 240;

    uint8_t buffer[screenHeight][screenWidth][3];
};

answered Mar 05 '20 at 12:07

Adrian Mole

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ok, my bad, but why doesn't it work anyway? shouldn't it truncate the value and use that as constant? the compiler error message is obscure to say the least – Luca Mar 05 '20 at 12:10
it's a zero, that's why – Luca Mar 05 '20 at 12:11
It does 'truncate' - but to zero! (You will get the same error message if you declare either of the literals as `0`.) – Adrian Mole Mar 05 '20 at 12:12

score 2 · Answer 3 · answered Mar 05 '20 at 12:09

2

Your issue is with uint8_t

static const uint8_t screenWidth = 256;//effectively 0

overflows and is exactly a big round ZERO. See integer-overflow.

To fix, switch to eg. size_t (also more appropriate for sizes)

answered Mar 05 '20 at 12:09

darune

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you link integer-overflow but this is unsigned integer overflow, one is (still) UB the other not – 463035818_is_not_an_ai Mar 05 '20 at 12:18
@idclev463035818 that tag is for both i believe. – darune Mar 05 '20 at 12:23
yes, but here the distinction matters. Not a big deal, just thought it is worth mentioning – 463035818_is_not_an_ai Mar 05 '20 at 12:23
@idclev463035818 It is only UB in arithmetic operations. Conversion from integral type to signed integral type is modulo 2^N in C++20 and was *implementation-defined* before. Calling it "*overflow*" at all is probably inaccurate. – walnut Mar 05 '20 at 14:46
@walnut thanks for clarification. Learned something today: check ;) – 463035818_is_not_an_ai Mar 05 '20 at 15:07

Initializing a two dimensional array with class static const integers

3 Answers3