Thread: help on test quest

I'll give you two hints...

1. an array name is actually a pointer to the first position in the array

2. A multidimensional array is in reality an array of arrays. And to dig in even further, a multidimensional array is in fact a single dimension array. arr[2][3] is in fact an array with 6 elements

Originally Posted by Mario F.

And to dig in even further, a multidimensional array is in fact a single dimension array. arr[2][3] is in fact an array with 6 elements

Really? Does arr[1][1] equal ((int*)arr)[4]?

A multidimensional array isn't usually a single dimension array. Consider argv. Each argument is a different length.

Yes. I should have rephrased that better. I meant conceptually since,

Code:

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

//and

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

produce the same array.

The second definition helps making it clear which answer is the right one.

arr[2][3], and arr[6] produce essentially the same array. In memory there is actually no difference. C++ strictly speaking doesn't know multi-dimension arrays.

Besides b) is wrong

almost

*(&a[0][0] + 4*i + j -1)

Which makes it a too complicated formula.

You are correct. a) is right. All others are wrong

arr[2][3], and arr[6] produce essentially the same array. In memory there is actually no difference. C++ strictly speaking doesn't know multi-dimension arrays.

I think it is more like the computer does not know about multi-dimensional arrays, and C++ initializer syntax allows one to take advantage of that as a shorthand. C++ does know about multi-dimensional arrays.

Yes. Try initializing your variables first.

Also, it looks like you're using an out-of-date compiler (if that actually compiles)

Code:

#include <iostream>

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

int test();

int main(void)
{
      std::cout<<test();
      std::cin.get();

      return 0;
}

int test()
{
if (a[i][j]==*(&a[0][0] + 4*i + j))
  return 1;

return 0;
}

almost

*(&a[0][0] + 4*i + j -1)

Which makes it a too complicated formula.

hmm... suppose i = j = 0.
*(&a[0][0] + 4*0 + 0 - 1) => *(&a[0][0] - 1)

This looks like one is trying to dereference a pointer that points to one before the first element.

Right away, this is dereferencing a memory address, which will make the program explode. It is wrong.

I think it is okay. This is from section 5.3.1 of the C++ Standard:
"The unary * operator performs indirection: the expression to which it is applied shall be a pointer to an object type, or a pointer to a function type and the result is an lvalue referring to the object or function to which the expression points. If the type of the expression is "pointer to T," the type of the result is "T.""
and a few lines later:
"The result of the unary & operator is a pointer to its operand. The operand shall be an lvalue or a qualifiedid. In the first case, if the type of the expression is "T," the type of the result is "pointer to T.""

In this case a[0][0] is an lvalue, so &a[0][0] is valid. (&a[0][0] + 4*i + j) is just pointer arithmetic, and results in a pointer. Hence using unary operator* on it should be valid, assuming that the pointer is valid.