What does this actually mean?

[ziwu]

int* ptrArray[ARRAY_SIZE];

Does it mean: define an array of pointers pointing to data type of int?

Does it differ from "int **matrix;" by the fact that both dimentions of
"matrix" is unknown but one dimention of "ptrArray" is known at compile
time?

Thanks for your help!

 

[Lew Pitcher]

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int* ptrArray[ARRAY_SIZE];

Does it mean: define an array of pointers pointing to data type of int?

It means that ptrArray is defined as an ARRAY_SIZE array of pointers to integers.

Does it differ from "int **matrix;"

Yes.

In this case, matrix is defined as a pointer to a pointer to an integer. That
is definitely not the same thing as an ARRAY_SIZE array of pointers to integers.

by the fact that both dimentions of
"matrix" is unknown but one dimention of "ptrArray" is known at compile
time?

Partially. But more completely, matrix is different from ptrArray because it
is a pointer, while ptrArray is an array.

- --
Lew Pitcher

Master Codewright & JOAT-in-training | GPG public key available on request
Registered Linux User #112576 (http://counter.li.org/)
Slackware - Because I know what I'm doing.
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[William L. Bahn]

int* ptrArray[ARRAY_SIZE];

Does it mean: define an array of pointers pointing to data type of int?

Yes. And most importantly (wrt to your next question) memory is allocated to
hold the specified number of pointers. The identifier ptrArray is a pointer
to the beginning of this space.

Does it differ from "int **matrix;" by the fact that both dimentions of
"matrix" is unknown but one dimention of "ptrArray" is known at compile
time?

int **matrix;

This only allocates enough memory to store a single pointer. It is
unitialized and no memory has been been alocated to store any int pointers.