CODEWITHSUNDEEP
c++arraysstruct
Xuhang
Xuhang

Reputation: 495

initialize std::array <struct, size>

I am trying to defined and initialize an array of struct.

#include <iostream>
#include <array>

int main() {
    struct row{
        double a0;
        double a1;
    };

    //method 0: this way works
    row c[2] ={{1.0,2.0},{3.0,4.0}};

    //method 1: declare and initialization in same line 
    //std::array<row, 2> a = { {1.0, 2.0}, {3.0, 4.0} };//error: Excess elements in struct initializer
    std::array<row, 2> a = {{ {1.0, 2.0}, {3.0, 4.0} }}; //double brace


    //method 2, declare, then initialize in different line
    std::array<row, 2> b;
    //b = { {1.0, 2.0}, {3.0, 4.0} };//error: No viable overloaded '='
    b = { { {1.0, 2.0}, {3.0, 4.0} } }; //double brace

    return 0;
}

Now I find double brace works from this post.

Just wondering why do we need extra pair of brace to initialize array of struct?

Upvotes: 2

Views: 1178

Answers (1)

Adrian Mole
Adrian Mole

Reputation: 51825

The literals (without the doubled braces) you are trying to use to initialize/assign your std::array variables do not match the type of those arrays. You need to explicitly make each of the 'top-level' elements a row object, like this, for example:

int main()
{
    struct row {
        double a0;
        double a1;
    };
    
    std::array<row, 2> a = { row{1.0, 2.0}, row{3.0, 4.0} };

    std::array<row, 2> b;
    b = { row{1.0, 2.0}, row{3.0, 4.0} };

    return 0;
}

This is because, without the double braces, your RHS literals are expected to be objects of the std::array<row,2> class (unambiguously). However, with the double-braces, you are using aggregate initialization rather than (copy) assignment (as mentioned in the post you link).

Upvotes: 4

Related Questions