Why can't you pass 2d pointers as const?

[neseidus]

I don't understand why the following doesn't compile:

#include <iostream>
using namespace std;

/**
 *	Utility function used to allocate 2-D arrays
 */
template <typename T>
T** new2DArray( const int& d1, const int& d2 )
{
	T ** result = new T*[d1];
	result[0] = new T[d1 * d2];

	for (int row=1; row < d1; row++)
	{
		result[row] = result[row-1] + d2;
	}

	return result;
}

void t1( const int* test )
{
     // Can't modify array
}

void t2( const int** test )
{
     // Can't modify 2D array
}

void main()
{
	int* test1 = new int();
	int** test2 = new2DArray<int> (3, 3);

	t1( test1 );	// Compiles
	t2( test2 );	// Doesn't compile

	delete test1;

	delete test2[0];
	delete test2;
}

 

[ArkM]

It's a correct compiler behaviour. No legal conversion

T** => const T** (and vice versa)

in C++.
There are two legal conversions:

T  => const T
T* => const T*

Conversion "T** --> const T**" cannot be derived from those two rules (MSDN, article Q87020, for example).

Don't uncheck Process TGML flag in the form...

You can force a conversion by brutal force method cast...

 

[neseidus]

hmm Alright thanks

 

[PerFnurt]

You can encapsulate the arrays to solve the problem.

#include <vector>

typedef std::vector< std::vector <int> > My2DArray;

...
void someFunc(const My2DArray& arr)
{
  ...
}

/Per
[sub]
"It was a work of art, flawless, sublime. A triumph equaled only by its monumental failure."[/sub]