Factor function passed with reference

Question

Good afternoon gentlemen

I need to make an algorithm with a procedure that takes a variable by reference and then updates the value of the variable to its corresponding factorial.

That’s the algorithm I made:

#include<stdio.h>
int fatorial(int *n){
    if ((*n==1) || (*n==0)) return 1;               
    else return fatorial(*n-1)*n;
}
int main(){
    int n = 5;
    printf("%d\n",fatorial (&n));
}

But you’re making a mistake on the line:

else return fatorial(*n-1)*n;

Error:

[Warning] passing argument 1 of 'fatorial' makes pointer from integer without a cast
[Note] expected 'int *' but argument is of type 'int'
[Error] invalid operands to binary * (have 'int' and 'int *')

Can someone help me solve this problem?

Answer 1

In your question you have very well described just what you have not implemented:

... receive a variable by reference and then update the variable value...

If you are receiving by reference and need to change the reference itself, the function will not return the result, but must assign the result to the pointer itself.

void fatorial(int* n) {
    int temp, resultado = 1;

    if (*n > 1) {
        temp = *n - 1;
        fatorial(&temp);
        resultado = (*n) * temp;
    }

    *n = resultado;
}

So, when doing, for example:

int n = 5;
fatorial(&n);

printf("%d", n);

It will be displayed the value 120, referring to the value of 5! calculated.

The idea of the code is:

1. Defines the variables temp, which will store the result of fatorial(n-1) if necessary, and resultado, which will store the final result;
2. If the value of n is greater than 1, recursively calculates the factorial;
3. Sets the value of temp as being the value of n down by 1;
4. Calls the function of fatorial passing the reference of temp, thus temp will have the factor value of n-1;
5. Sets the result value as the value of n multiplied by the value of temp;
6. Updates the reference value to the resultado;

Answer 2

#include<stdio.h>
int fact(int *a,int *p) {
    while(*a>1){
            *p*=*a;
            *a=*a-1;
            }
    }
 int main() {
        int num, factorial=1,c,*p;
        p=&factorial;
        printf("entre um numero:");
        scanf("%d",&num);
        fact(&num,p);
        printf("fatorial do numero é:%d",*p);

        return 0;
        }

Answer 3

Implementation of factorial in C:

int factorial(int n)
{
    if (n >= 1)
        return n*factorial(n-1);
    else
        return 1;
}

The error you have is because you are passing an int parameter and you need to pass an int reference to what you had previously

int *n

Passing value by reference using a loop:

void factorial(int *f){
    int i, temp = 1;
    for (i = 1; i <= *f; i++)
        temp = temp * i;
    *f = temp;
}