difference between (int) malloc(nsizeof(int)) and int array = malloc(sizeof(int)n)

Asked Feb 03 '20 at 21:00

Active Nov 08 '21 at 14:28

Viewed 38 times

I'm doing an exercise where I'm allocating my array data to the heap instead of the stack.

so I changed my code from:

int array[n];

int *array = (int*) malloc(n*sizeof(int));

When I'm looking at my friends code that got the same exercise, they wrote int *array = malloc(sizeof(int)*n);

so, what is the difference between int *array = (int*) malloc(n*sizeof(int)); and int *array = malloc(sizeof(int)*n); ?

edited Nov 08 '21 at 14:28

asked Feb 03 '20 at 21:00

Theirs is good without the cast, and some would prefer it like this: `int *array = malloc(n * sizeof *array);` – Weather Vane Feb 03 '20 at 21:04
okey so its the same thing and both is working? is it like in math that 10/4 is the same as 5/2 ? – Feb 03 '20 at 21:13
Have you read the linked question? – Weather Vane Feb 03 '20 at 21:14
2

Better to use `int *array = malloc(sizeof *array * n);` and avoid trying to get and maintain the right type. – chux - Reinstate Monica Feb 03 '20 at 21:57

difference between (int*) malloc(n*sizeof(int)) and int *array = malloc(sizeof(int)*n)