Initialize a two-dimensional std::array of type enum class (C++11)

Question

I have the following class in C++11:

class MyTable
{
    public:
    enum class EntryType
    {
        USED, FREE
    };

    MyTable(EntryType value)
    {
        for (uint32_t i = 0; i < 10; ++i)
        {
            memset(_table[i].data(), (int)value, sizeof(_table[0][0]) * 50);
        }
    }
    array<array<EntryType, 50>, 10> _table;
}

Trying to construct an object of MyTable with value of EntryType::FREE, every item in the 2-dimensional array has the value of 0x01010101 (1b every 8 bits), instead of the expected value of just 0x1

I'm guessing it has something to do with my value being casted to int, but I don't know what I'm supposed to do in order to fix it.

Why do you use memset to initialize the values at all, instead of directly assigning them? — πάντα ῥεῖ, May 26 '17 at 09:09
That's how `memset` is supposed to work, it initializes byte-wise. — Kinan Al Sarmini, May 26 '17 at 09:11
Just use manual initialization. If you will try to set `1` to `int` type via `memset()` is will be `0x01010101` and it's defined behaviour — renzo, May 26 '17 at 09:16
Your implementation is not very memory efficient, if you do not need 32bit enum you can use an array of uint8_t, and then your memset() call will succeeed. — gabry, May 26 '17 at 10:48
@gabry you are right, but memory is not an issue for my current task. — itzhaki, May 26 '17 at 12:23
@renzo Implementation-defined behaviour, you mean? The `sizeof(int)` is not necessarily 16 bits as that hex pattern implies. — underscore_d, May 26 '17 at 21:52

gsamaras · Accepted Answer · 2017-05-26T09:27:56.557

6

memset() is expected to work that way, since it's

sets each byte of the destination buffer to the specified value.

Read more in Why is memset() incorrectly initializing int?

However, be careful, since as juanchopanza said, std::array may have padding at the end (read more in std::array alignment ), which means that this approach might fail.

Since it's a 2D array, you could use a range-based-for loop and std::array::fill, like this:

for(auto& row : _table)
    row.fill(value);

as renzo stated.

In case you do not want to set every row at the same value, you could do it like this of course:

for(auto &row : array)
    for(auto &col : row)
         col = value;

Read more in range-based for on multi-dimensional array.

edited May 26 '17 at 09:27

answered May 26 '17 at 09:11

gsamaras

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1

range-based-for could also be used. – The Techel May 26 '17 at 09:14
Or another variant: `for(auto& array : _table) array.fill(value);` – renzo May 26 '17 at 09:20
1

`std::array` can have padding at the end, so `memset` isn't guaranteed to work when used as in OP's code. – juanchopanza May 26 '17 at 09:22
@juanchopanza ah thank you, didn't know that! Thank you The Techel and renzo for the suggestions, they really improved the answer! =) – gsamaras May 26 '17 at 09:27
1

That solved my problem, thank you for the detailed reply. The `int` in `memset` threw me off, as I was not aware it is casted to `unsigned char` – itzhaki May 26 '17 at 10:05
@juanchopanza (notwithstanding how the implementation putting padding there seems stupid to me) Why would padding at the end make any difference to this? The `memset` would be performed from the beginning until the last element, inclusive. The padding at the end, after the last element, should not be touched nor read, so it shouldn't be relevant at all. – underscore_d May 26 '17 at 21:56

score 3 · Answer 2 · answered May 26 '17 at 09:19

3

This can be done with a ranged-based for loop and the std::array::fill member function.

MyTable(EntryType value)
{
    for (auto& row : _table) {
        row.fill(value);
    }
}

This will continue to work even if you change the array dimensions.

answered May 26 '17 at 09:19

Blastfurnace

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Initialize a two-dimensional std::array of type enum class (C++11)

2 Answers2