invalid conversion from *void to *int [-fpermissive] using malloc(sizeof()) - c

Question

I'm writing a program that calculates the greatest common denominator of two numbers, but i'm getting problem with malloc function and pointers. Actually it's clear how the stack and the heap segments work in the memory and why. But yet i'm not yet able to understand when declaring a pointer and using malloc is functional or not, is necessary or not, in a program. here is the code :
#include <iostream>
#include <stdlib.h>
#include <stdio.h>
int *calcolaDivisori(int);
int main(int argc, char** argv) {
int foundCounter = 0;
int i,j,s1,s2;
int n1,n2;
int mcd = 1,mcm;
int *pn1,*pn2;
int d1[100],d2[100];
// INPUT dei due interi
printf("Inserisci il primo numero :");
scanf(" %d", &n1);
printf("\nInserisci il secondo numero :");
scanf(" %d", &n2);
// calcolo divisori del primo e del secondo numero e li assegno ai relativi array
pn1 = calcolaDivisori(n1);
if (!pn1) return 1;
pn2 = calcolaDivisori(n2);
if (!pn2) return 1;
for (i=0;i<n1;i++) {
d1[i] = pn1[i];
}
for (i=0;i<n2;i++) {
d2[i] = pn2[i];
}
free(pn1);
free(pn2);
// confronto i divisori e calcolo il MCD
s1 = sizeof(d1) / sizeof(int);
s2 = sizeof(d2) / sizeof(int);
for(i=0; i<s1; i++) {
for (j=foundCounter; j<s2;j++) {
if (d1[i] == d2[j]) {
mcd*= d1[1];
foundCounter = j+1;
break;
}
}
}
printf("\n\nIl minimo comune divisore e' : %d", mcd);
return 0;
}
int *calcolaDivisori(int num) {
int i;
int *a = malloc(num * sizeof(int));
if (!a) return NULL;
for (i=2;i<num;i++) {
if (num%i == 0) {
num/=i;
a[i-2]=i;
}
}
return a;
}
I get the error in the title when is run the command :
int *a = malloc(sizeof(int));

Answer 1

You need to cast:
int *a = (int*)malloc(num * sizeof(int));
Because there's no implicit conversion from void* to type * in C++.
Note that this cast is not required in C and could potentially be dangerous to do so in C.
Except for #include <iostream>, nothing in your code is C++. So remove it and compile it with a C compiler and you wouldn't need this cast.