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
– walnut, Commented Mar 5, 2020 at 14:38

Vlad from Moscow · Accepted Answer · 2020-03-05 12:05:06Z

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 5, 2020 at 12:05

Vlad from Moscow

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1 Comment

Vlad from Moscow Over a year ago

@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.

Adrian Mole · Accepted Answer · 2020-03-05 12:07:51Z

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 5, 2020 at 12:07

Adrian Mole

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3 Comments

Luca Over a year ago

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 Over a year ago

it's a zero, that's why

Adrian Mole Over a year ago

It does 'truncate' - but to zero! (You will get the same error message if you declare either of the literals as 0.)

darune · Accepted Answer · 2020-03-05 12:09:38Z

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 5, 2020 at 12:09

darune

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5 Comments

463035818_is_not_an_ai Over a year ago

you link integer-overflow but this is unsigned integer overflow, one is (still) UB the other not

darune Over a year ago

@idclev463035818 that tag is for both i believe.

463035818_is_not_an_ai Over a year ago

yes, but here the distinction matters. Not a big deal, just thought it is worth mentioning

walnut Over a year ago

@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.

463035818_is_not_an_ai Over a year ago

@walnut thanks for clarification. Learned something today: check ;)

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