Array initialization with structures

Question

I have the following data structures

struct single_t
{
    uint16_t i1 = 0;
    uint8_t i2 = 0;
    uint8_t i3 = 0;
};

struct mapping_t
{
    uint8_t n1;
    uint8_t n2;
    bool enable;
    uint n3;
    std::array<single_t, 32> map;
};

I want to initialize them in the following way:

mapping_t m1 {
    3,                                  // n1
    254,                                // n2
    true,                               // enable
    5,                                  // n3

    // map
    // i1                   i2              i3
    {{
        {0x1000,            1,              8}
    }}
};

can i be sure, that the elements in the std::array<single_t, 32> map;, in this case indexes 1..31, are initialized to 0 or is it like the uninitialized variable on the stack int i; ? My debugger shows they are at 0, but is that implementation dependent on the debug build or is this defined in the standard?

The elements 1 - 31 of m1 are value initialized, i.e. initialized to zero in your case. Your code is fine. — R Sahu
– R Sahu, Commented Feb 25, 2017 at 8:20

hivert · Accepted Answer · 2017-02-25 08:30:36Z

2

From http://en.cppreference.com/w/cpp/language/aggregate_initialization

If the number of initializer clauses is less than the number of members or initializer list is completely empty, the remaining members are value-initialized. If a member of a reference type is one of these remaining members, the program is ill-formed.

The default case for value initialization is to initialize with 0

See http://en.cppreference.com/w/cpp/language/value_initialization

The effects of value initialization are:

[...]

4) otherwise, the object is zero-initialized.

To sumarize, you're fine !

answered Feb 25, 2017 at 8:30

hivert

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