Passing 2-D array as argument

Question

I am trying to pass a 2-d array to a function which accept a pointer to pointer. And I have learnt that a 2-d array is nothing a pointer to pointer(pointer to 1-D array). I when I compile the below code I got this error.

#include<iostream>

void myFuntion(int **array)
{
}
int main()
{
   int array[][]= {{1,2,3,4},{5,6,7,8,9},{10,11,12,13}};
   myFuntion(array);
   return 0;
}

In function 'int main()': Line 5: error: declaration of 'array' as multidimensional array must have bounds for all dimensions except the first compilation terminated due to -Wfatal-errors.

Can anybody clear my doubt regarding this and some docs if possible for my more doubts.

Arrays are not pointers. It isn't a pointer to a pointer, and it decays to a pointer to an array when it's passed into a function. — Qaz
– Qaz, Commented Oct 20, 2012 at 16:46
array[][] must have some size defined. I guess we are not allowed to specify an array without any size specified. — Afaq
– Afaq, Commented Oct 20, 2012 at 16:49
@FatimaZohra, Actually, you need to specify the other dimensions, not the first. Anyway, that's a 2D array of int pointers, and OP, your sets need to all contain the same number of elements so that you can specify the second dimension. — Qaz
– Qaz, Commented Oct 20, 2012 at 16:51

vmp · Accepted Answer · 2012-10-20 20:59:54Z

8

  void myFunction(int arr[][4])

you can put any number in the first [] but the compiler will ignore it. When passing a vector as parameter you must specify all dimensions but the first one.

answered Oct 20, 2012 at 20:59

vmp

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md5 · Accepted Answer · 2012-10-20 16:59:33Z

1

You should at least specify the size of your second dimension.

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

There is also an error which is often repeated. To pass a 2D array as argument, you have to use the following types:

void myFuntion(int (*array)[SIZE2]);
/* or */
void myFuntion(int array[SIZE1][SIZE2]);

answered Oct 20, 2012 at 16:59

md5

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4 Comments

yogi Over a year ago

can you please give your review on this [link] codepad.org/n8GRm9dU [/link] Error:- error: cannot convert 'int ()[4]' to 'int*' for argument '1' to 'void myFuntion(int**)' why does this mean as I try to pass myFunction(array);

yogi Over a year ago

Can you give a logic behind this. void myFuntion(int (*array)[SIZE2]);

md5 Over a year ago

array is declared as a pointer to an array of size SIZE2. SIZE2 is here the size of your second dimension.

Raj Kumar Mishra Over a year ago

Actually these both are same ---> int (*array)[SIZE2] = int array[SIZE1][SIZE2]).......................................... array[SIZE1][SIZE2] = ( *(array + SIZE1)[SIZE2] ) = ( *(array + 0)[SIZE2]) = *array[SIZE2]. Here 1st argument is ignored by compiler hence SIZE1 = 0.

zabulus · Accepted Answer · 2012-10-20 17:02:42Z

1

Why don't use std::vector instead of "raw" arrays. Advantages:
1. It can dynamically grow.
2. There is no issues about passing arguments to the function. I.e. try to call void myFuntion(int array[SIZE1][SIZE2]); with array, that has some different sizes not SIZE1 and SIZE2

answered Oct 20, 2012 at 17:02

zabulus

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Comments

Matthieu Brucher · Accepted Answer · 2019-03-06 17:41:16Z

1

Another templated solution would be:

template<int M, int N>
void myFunction(int array[N][M])
{
}

answered Mar 6, 2019 at 17:41

Matthieu Brucher

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Comments

Fatima Zohra · Accepted Answer · 2012-10-20 17:02:55Z

0

#include<iostream>
 void myFuntion(int arr[3][4]);
int main()
  {
  int array[3][4]= {{1,2,3,4},{5,6,7,8},{10,11,12,13}};
 myFuntion(array);
  return 0;
 }
   void myFuntion(int arr[3][4])
   {

   }

http://liveworkspace.org/code/0ae51e7f931c39e4f54b1ca36441de4e

answered Oct 20, 2012 at 17:02

Fatima Zohra

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Omkant · Accepted Answer · 2012-10-20 17:10:03Z

0

declaration of ‘array’ as multidimensional array must have bounds for all dimensions except the first So you have to give

array[][size] //here you must to give size for 2nd or more

For passing the array in function , array is not a pointer to a pointer but it's pointer to an array so you write like this

fun(int (*array)[])

Here if you miss the parenthesis around (*array) then it will be an array of pointers because of precedence of operators [] has higher precedence to *

answered Oct 20, 2012 at 17:10

Omkant

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Comments

Satyansh · Accepted Answer · 2023-12-12 16:50:16Z

0

you need to pass array[][n] like this. you can leave 1st bracket but you have to put data into second.

answered Dec 12, 2023 at 16:50

Satyansh

1

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