CODEWITHSUNDEEP
c++arrayspointersarguments
nauti
nauti

Reputation: 1436

C++ pointer array argument

I have the following code:

void getPossibilities(int *rating[200][3]){
// do something
}

int main ()
{
int rating[200][3];
getPossibilities(&rating);
}

this throws following error message:

error: cannot convert int ()[200][3] to int ()[3] for argument 1 to void getPossibilities(int (*)[3])

Upvotes: 2

Views: 2163

Answers (3)

Sarfaraz Nawaz
Sarfaraz Nawaz

Reputation: 361254

The function signature should be this:

void getPossibilities(int (*rating)[3]);

and pass argument as:

getPossibilities(rating);

The variable rating is a two dimentional array of form T[M][N] which can decay into a type which is of form T(*)[N]. So I think that is all you want.

As in the above solution the array decays, losing the size of one dimension (in the function you only know N reliably, you just loss M due to the array-decay), so you've to change the signature of the function to avoid decaying of the array:

void getPossibilities(int (&rating)[200][3]) //note : &, 200, 3
{
  //your code
}

//pass argument as before
getPossibilities(rating);  //same as above

Better yet is to use template as:

 template<size_t M, size_t N>
 void getPossibilities(int (&rating)[M][N])
 {
       //you can use M and N here
       //for example
       for(size_t i = 0 ; i < M ; ++i)
       {
          for(size_t j = 0 ; j < N ; ++j)
          {
             //use rating[i][j]
          }    
       }
 }

To use this function, you've to pass the argument as before:

 getPossibilities(rating); //same as before!

Upvotes: 5

iammilind
iammilind

Reputation: 69958

When passing an array of N-dimensions to a function, the 0th dimension is always ignored. That's why a[N] decays to *p. In the same way, a[N][M] decays to (*p)[M].
Here, (*p)[M] is a pointer to an array of M elements.

int a1[N][M], a2[M];
int (*p)[M];
p = a1; // array a1[N][M] decays to a pointer
p = &a2; // p is a pointer to int[M]

So your function signature should be:

void getPossibilities(int (*rating)[3]);

Now since you are using, C++, it's worth taking advantage of its facility where you can pass an array by reference. So preferred way is:

void getPossibilities(int (&rating)[200][3]);

Upvotes: 0

Šimon T&#243;th
Šimon T&#243;th

Reputation: 36423

There is a difference between int (*x)[200][3] and int *x[200][3]

Upvotes: 0

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