Malloc

**incognito** · 02-24-2004

Hey guys I am taking a Data Structure class now, and we're using C, but I don't have much experience with C, and as so don't really remember how to use malloc, can anybody help me here. I know it's very similar to it's "new" counterpart in C++.

Code:


#include <stdio.h>

int main()

{

	int *Articulos;
	int Cantidad;

	printf("Cuantos articulos va a facturar?\n");
	scanf("%d",&Cantidad);

	Articulos=(*int)malloc(Cantidad * sizeof int);


	return 0;


}

It says something about "type int unexpected";

**Thantos** · 02-24-2004

In C the proper way is

Code:

Articulos = malloc (Cantidad * sizeof int);

**incognito** · 02-24-2004

I tried that and it still gave me the error. HOwever when I take out the sizeof int thing it works. (at least it does not give me any errors)

**Thantos** · 02-24-2004

try sizeof(int)

**c99** · 02-25-2004

incognito.

Articulos=(*int)malloc(Cantidad * sizeof int);

(*int) should be (int *)

IMO I'd #include stdib.h and not worry about the cast at all.

**Edward** · 02-25-2004

You shouldn't cast the return of malloc in C. It hides the error of not including stdlib.h. An accepted convention of calling malloc is to multiply the amount by the size if the dereferenced pointer that you're assigning to, then return without a cast.

Code:

Articulos = malloc(Cantidad * sizeof *Articulos);

This works even though Articulos is an uninitialized pointer because sizeof doesn't evaluate the expression you give it. It only returns the final type of that expression, which in this case would be int.

Also, sizeof doesn't require parentheses around the expression unless it is a type name. So sizeof int is ill-formed. It should be sizeof(int).

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