question about malloc

[mouse666666]

hi i have this program i wanted to know what int* is doing. i know that if i enter a number. it will be (y*int) for example if i enter 4 it is. 4X4=16. because int has 4 bytes and entering 4 marks. int has 4 bytes.

so my question is what is int* doing. also how can i printed out the size of the dynamic memorry

#include<stdio.h>
#include<conio.h>
#include<stdlib.h>

main()
{
      int *marks;
      int y
      int t;
      
      printf("enter the mark);
      scanf("%i",&y);
      marks=(int*)malloc(y*sizeof(int)) //what is int * doing  why does it equals to mark

}

[Dino]

In that context, (int *) is a cast. It it casting the pointer returned by malloc as a pointer to an int.

[mouse666666]

so u are saying that it is return (int*) as a pointer if i understood you. and the marks will point to int* right. how can i print the total bytes of the dynamic array

[Elysia]

int* is a type. You cannot point to a type. It points to whatever value malloc returns whose type is void*. The cast changes the type of the returned value from void* to int* (unnecessary in C).
The only way to print the size of the allocation is to remember how much you've allocated. There is no standard method in C to get the size of an existing allocation.

[Salem]

> marks=(int*)malloc(y*sizeof(int)) //what is int * doing why does it equals to mark
In the context of a valid C program, compiled with a C compiler - absolutely nothing.

It's a waste of key strokes.

There's a FAQ on (NOT) casting the return result of malloc.