Put array variable in multidimensional array

Question

I want the following

int ring1[5] = {1,2,3,4,5};
int ring2[5] = {6,7,8,9,10};
int ring3[5] = {11,12,13,14,15};
int rings[3][5] = {ring1, ring2, ring3};

But I get an error on the last line:

Array Must be initialized with a brace-enclosed initializer.

Time to open up your text-book on the chapters on pointers. — Some programmer dude
– Some programmer dude, Commented Dec 10, 2020 at 11:25
Arrays cannot be copied by assignment. They can be copied with memcpy or element by element in a loop. You could make rings into an array of pointers, e.g. int *rings[3] = {ring1, ring2, ring3}; but note that any changes to ring1[i] will also change rings[0][i] and vice versa. Likewise for ring2 and ring3. — user3386109
– user3386109, Commented Dec 10, 2020 at 11:27
The error is kind of misleading , are you compiling using C++ compiler? — IrAM
– IrAM, Commented Dec 10, 2020 at 11:42

IrAM · Accepted Answer · 2020-12-10 11:57:13Z

2

In C name of the array indicates the base address of the array, so all of your ring1, ring2 and ring3 gives the address of their starting element (i.e 0th).

Similarly rings is also an array (Its a 2 Dimensional array).

We cannot store address in arrays, for that will have to use pointers.

So we have to use below way.

#include <stdio.h>

int main()
{
    int ring1[5] = {1,2,3,4,5};
    int ring2[5] = {6,7,8,9,10};
    int ring3[5] = {11,12,13,14,15};
    //int rings[3][5] = {{ring1}, {ring2}, {ring3}};
    int *rings[3] = {ring1, ring2, ring3};
    
    printf("ring1 = %p and rings[0] =%p\n", ring1, rings[0]);
    printf("ring2 = %p and rings[1] =%p\n", ring1, rings[1]);
    printf("ring3 = %p and rings[2] =%p\n", ring1, rings[2]);

    return 0;
}

edited Dec 10, 2020 at 11:57

answered Dec 10, 2020 at 11:31

IrAM

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1 Comment

underscore_d Over a year ago

Yeah, but this loses the size info for the 'nested' arrays, and other issues from using pointers. A less direct but more correct fix would be to iterate and assign the sub-arrays IMO

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