Thread: hello all quick pointers question

  1. #1
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    hello all quick pointers question

    i had to write this func. on an assignment

    Code:
    int checkMat(int**mat , int rows)
    {
    	int i,j;
    	int sum=0;    // will be the sum of the first - the sum of the second if sum=0 then they're equal
    	for (i=0;i<rows;i++)
    		for(j=0;j<rows;j++)
    			sum = sum+mat[i][j]-mat[rows-i][rows-j];
    	if (sum=0) return 1;
    	return 0;
    
    }
    now i want to test it in the main so i did :
    Code:
    	int arr[2][2]={{3,5},
    					{9,4}};
            printf("%d",checkMat(arr[0],6));
    and it gives a "argument of type incompatible" what am i doing wrong in the main ?

    i don't want to change the signature of the function
    tnx all helpers !

  2. #2
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    Your function wants int **mat... you are handing it an array.

  3. #3
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    so i have to send it a 2d array pointer ? i can't send the array like in regulars arrays ?

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  5. #5
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    ok i read it i understand why it doesn't work , so i tried to make a pointer to point to the 2d array like so:
    Code:
    	int arr[2][2]={{4,6},{3,6}};
    	int **p ;
    	**p[2]=&arr[0][0];
    i tried other variations in order to make it work , any ideas on how to make a pointer point to the 2d array ?

  6. #6
    and the hat of int overfl Salem's Avatar
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    Code:
    void foo ( int (*arr)[2] ) {
       // do stuff with arr[x][y]
    }
    int main ( ) {
      int arr[2][2]={{4,6},{3,6}};  // 2D array 
      int (*pa)[2] = arr;  // pointer to 2D array (or more accurately, pointer to the first row of the array)
      foo( arr );  // call it
      foo( pa );  // call it again
      return 0;
    }
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.

  7. #7
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    yea ok but my function signature is
    Code:
    int checkMat(int**mat , int rows)
    so i can't do
    Code:
    checkMat(arr)

  8. #8
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    If this is a two-dimensional array, shouldn't you be passing both the row count and the column count to the function? Actually, I guess if it's square it doesn't really matter. I'm just really specific about that sort of stuff.

  9. #9
    and the hat of int overfl Salem's Avatar
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    Well if you have int**, and you can't change it, then you can't pass a true 2D array to it without some additional code.

    Say
    Code:
    int arr[2][2]={{4,6},{3,6}};  // 2D array
    int *pa[2] = { &arr[0][0], &arr[1][0] };
    checkMat( pa, 2 );
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.

  10. #10
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    yea i forgot to say it was a square
    ok it works now thanks a lot
    can you just explain a little about what you did there ?
    is it like an array of pointers ?

  11. #11
    and the hat of int overfl Salem's Avatar
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    int (*pa)[2]
    Pointer to an array.

    int *pa[2]
    Array of pointers.
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.

  12. #12
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    oh i c , thanks a lot
    it really cleared some issues for me

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