Difference between double** and &arr, where arr is a double[]

Asked Jun 04 '13 at 19:52

Active Jun 04 '13 at 19:55

Viewed 87 times

I am using C. I have a double array like this:

double arr[3]={1,2,3};

Next, I assumed that a

double[]

is just like

double *,

and thus I created this pointer variable:

double ** ppdArr = &arr; /* indirection */

I planned to dereference it later and access the items in the original array as follows:

double* pdArr = *ppdArr; /* dereference */
double value = pdArr[0]; /* access violation, because pdArr is NULL */

Why? When using arrays and pointers interchangeably, what is the difference between static vs. dynamic allocation? Or are there other reasons for this error? For example, the following works (as expected):

double *arr2 = ... /*allocated through malloc()*/
double ** ppdArr2 = &arr2; /* indirection */
double *pdArr2 = *ppdArr2; /* dereference */
double value2 = pdArr2[0]; /* works */

asked Jun 04 '13 at 19:52

Saad Quader

I am using Visual Studio 2008. – Saad Quader Jun 04 '13 at 19:53
2

`double ** ppdArr = &arr;` shouldn't compile. Post something we can compile? – Pubby Jun 04 '13 at 19:54
Uhm... how can code that doesn't compile crash? – Nik Bougalis Jun 04 '13 at 19:58
The point of confusion is that you can use array notation on pointers, even though functionally they are quite different, especially when you get into multiple dimensions. – Hot Licks Jun 04 '13 at 19:59
@Pubby @NikBougalis: The code `double ** ppdArr = &arr;` compiles. Here is the code in ideone (gcc-4.8.1): [link](http://ideone.com/oT9vKR): it compiles, but makes runtime error. – Saad Quader Apr 04 '14 at 14:32

Difference between double** and &arr, where arr is a double[]

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