Initializing Array of Arrays - Trouble

Question

OK, I know that in C++ a - let's say 2-dimensional - array can be initialized this way :

int theArray[5][3] = { 
    {1,2,3},
    {4,5,6},
    {7,8,9},
    {10,11,12},
    {13,14,15} 
};

Now, what if I want to use pre-existing arrays as theArray's elements?

E.g.

// A, B, C, D,... have already been declared as :
// `const U64 A[] = { 1,2,3,4 };` etc...

const U64 multiDimArray[12][64] = { 
     A, B, C, D, E, F,  
     G, H, I, J, K, L
};

This one, throws an error though :

cannot initialize an array element of type 'const U64' 
(aka 'const unsigned long long') with an lvalue of type 'const U64 [64]'

I see the point, but hopefully you can see mine.

Is there a workaround so that I can easily achieve the same thing? (Any suggestion - perhaps something using Boost? - is welcome)

Pavel Shishpor · Accepted Answer · 2023-10-06 11:23:38Z

6

If you use C++11, the initializer list for an array is flexible:

std::array<std::array<U64, 64>, 12> multiDimArray = {
     A, B, C, D, E, F,  
     G, H, I, J, K, L
};

will work fine, assuming A..L are std::array<64, U64>.

The array does have no overhead to the c-style array. Click here for official reference.

"The size and efficiency of array<T,N> for some number of elements is equivalent to size and efficiency of the corresponding C-style array T[N]." (From the reference)

I said "flexible", since you can use a mixed initializer-list like this:

std::array<int, 3> first_row = {1,2,3};
std::array<std::array<int, 3>, 2> a={
  first_row,
  {2, 2, 2}
};

You can use this as a fixed-size array, with the same operations:

a[1][2]=2; a[0][1]=1; a[0][2]=3;

edited Oct 6, 2023 at 11:23

Pavel Shishpor

1,02711 silver badges15 bronze badges

answered Dec 15, 2012 at 8:06

Barney Szabolcs

12.7k16 gold badges71 silver badges97 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Stephen Chu Over a year ago

And you can use the new array class for fixed size array.

TimoteeTheCodeMonkee · Accepted Answer · 2012-12-15 07:51:50Z

5

I can see why this is useful, however, in C, using just the variable name of an array will return the address of the array in memory. The compiler has no idea as to what is actually stored at A during compile time, so this wouldn't work.

As an alternative, what you could do is either use a memcpy and copy the elements into the array (but then it won't be const), or you could use a #define A { 1, 2, 3, 4 } and then maybe do something like:

#define A_val { 1, 2, 3, 4 }
#define B_val { 5, 6, 7, 8 }
const U64 multiDimArray[12][64] = {
    A_val,
    B_val,
    // and so on and so forth
}
const U64 A[4] = A_val; // if you need this
const U64 B[4] = B_val; // you can do it like this

answered Dec 15, 2012 at 7:51

TimoteeTheCodeMonkee

3321 silver badge5 bronze badges

1 Comment

Dr.Kameleon Over a year ago

This sounds like a really interesting idea. Let me see how I could fit this into my general code scheme. ;-)

iagreen · Accepted Answer · 2012-12-15 08:08:46Z

2

You can use the fact that a two dimensional array is actually a one dimensional array of pointers to your advantage here. The follow initialization should work for you.

const U64 *multiDimArray[12] = { 
     A, B, C, D, E, F,  
     G, H, I, J, K, L
};

answered Dec 15, 2012 at 8:08

iagreen

32.1k8 gold badges79 silver badges90 bronze badges

1 Comment

HolyBlackCat Over a year ago

"actually a one dimensional array of pointers" They aren't. Yours is, but yours isn't a true 2D array.

Collectives™ on Stack Overflow

Initializing Array of Arrays - Trouble

3 Answers 3

1 Comment

1 Comment

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

1 Comment

1 Comment

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related