CODEWITHSUNDEEP
carrayspointers
user3602824
user3602824

Reputation: 23

Pointers and Arrays in C language

I am by no means a master of the C language, nor am i going to claim to be. But I was pretty sure I understood pointers until I came across these two very different implementations that the author claimed would have the same outcome. Im just not seeing it. Can somebody please explain to me what each of these are doing?

int (*AA)[4];
AA = malloc(sizeof(int[4])*size);

and the 2nd one:

int *BB;
BB = (int *)malloc(size*sizeof(int));

From my current understanding, If i wanted to make the 2nd one in c++, it is equivalent to: 

int *CC;
CC=new int[size]

Is this assumption Correct?

Upvotes: 1

Views: 137

Answers (2)

ryyker
ryyker

Reputation: 23236

comments notwithstanding, Given your code:

enter image description here

For illustration, here is what is shown when in debug mode looking at what was created: (ANSI C99 compiler, 32bit build)

At the very least, understanding fully what your author meant when stating same outcome is important.
Also illustrates why the use of calloc() as opposed to malloc() might be a good option (C99).

The second part question: C++ version

int* BB = new int[size];

is valid.

Upvotes: 0

fstd
fstd

Reputation: 574

First part: int (*AA)[4]; defines AA to be a pointer to an int[4].

AA = malloc(sizeof(int[4])*size); allocates storage for size int[4]s

Second part: int *BB; defines BB to be a pointer to an int.

BB = (int *)malloc(size*sizeof(int)); allocates storage for size ints, then unnecessarily casts the result, thereby introducing a potential error you won't get warned about (namely, if the prototype for malloc() is not in scope).

Therefore, the result is arguably different

Upvotes: 3

Related Questions