Passing multidimensional array to a function (c++)?

Question

I've been trying to pass a multidimensional array, of an unknown size, to a function, and so far have had no luck, when the array is declared, its dimensions are variables:

double a[b][b];

As far as I can tell, I need to give the value of b when I declare the function, a can be unknown. I tried declaring b as a global variable, but it then says that it must be a constant.

ie:

int b;

double myfunction(array[][b])
{
}

int main()
{
int a;
double c;
double myarray[a][b];

c=myfunction(myarray);

return 0;
}

Is there any way get this to work?

Not very pretty, but can you not just pass in the pointer to the first element? — Chris
– Chris, Commented Jul 25, 2012 at 17:33
If the dimensions are variable use either std::vector or boost::multiarray. — R. Martinho Fernandes
– R. Martinho Fernandes, Commented Jul 25, 2012 at 17:34
I know this doesn't answer your question, but you are missing double in front of your array parameter. I don't know if this is an oversight in your post here or if it is missing from the code you are compiling as well. — Code-Apprentice
– Code-Apprentice, Commented Jul 25, 2012 at 17:46
@chris, std::vector makes life easier for single-dimension arrays but it complicates multi-dimension arrays since the size of each row must be set separately. — Mark Ransom
– Mark Ransom, Commented Jul 25, 2012 at 18:04

Mr.Anubis · Accepted Answer · 2012-07-25 21:07:54Z

4

Pass by value :

double myfunction(double (*array)[b]) // you still need to tell b

Pass by ref :

double myfunction(int (&myarray)[a][b]); // you still need to tell a and b

Template way :

template<int a, int b> double myfunction(int (&myarray)[a][b]); // auto deduction

edited Jul 25, 2012 at 21:07

answered Jul 25, 2012 at 17:44

Mr.Anubis

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Comments

Sean McCully · Accepted Answer · 2012-07-25 18:22:08Z

1

Perhaps reading some references on C++ and arrays would help,

http://en.cppreference.com/w/cpp/container/array

edited Jul 25, 2012 at 18:22

answered Jul 25, 2012 at 17:37

Sean McCully

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Comments

shotex · Accepted Answer · 2014-04-03 17:22:28Z

1

if you want to pass an array of unknown size you can declare an array in Heap like this

//Create your pointer
int **p;
//Assign first dimension
p = new int*[N];
//Assign second dimension
for(int i = 0; i < N; i++)
p[i] = new int[M];


 than you can declare a function like that: 
double myFunc (**array);

answered Apr 3, 2014 at 17:22

shotex

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Comments

Caesar · Accepted Answer · 2012-07-25 17:36:21Z

-1

void procedure (int myarray[][3][4])

2 Comments

John Humphreys Over a year ago

I think this is a 3-dimensional array instead of 2, or else you've found a syntax that I've never seen before [which is admittedly possible! :)]

Qaz Over a year ago

See Griwes's comment on the other answer. A better link would be this question.

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