在C中将指针转换为int

Question

I have to learn it for my study. Is their any way to cast a pointer to integer. I have to give myEulerForward1 a pointer as a paramter and i always get this error message :

eulerZahl.c: In function ‘main’:
eulerZahl.c:38:35: warning: passing argument 1 of ‘myEulerForward1’ makes pointer from integer without a cast [-Wint-conversion]
  double forward = myEulerForward1(k);
                                   ^
eulerZahl.c:16:8: note: expected ‘int *’ but argument is of type ‘int’
 double myEulerForward1(int *n1){
        ^~~~~~~~~~~~~~~

Can someone help me with it?

#include <stdio.h>
#include <stdint.h>


double kFactorial(int k){


    if(k <= 1){
        return 1;
    }

    return k * kFactorial(k - 1);
}


double myEulerForward1(int *n1){


    double n = 1;   
    double euler, nFact = 1;
    for(int i = sizeof(&n1); i >= 0; i--){
        nFact*= i;
        euler = euler + (1.0/nFact);
    }

    return euler;

}


int main(){

    int k = 4;
    double factorial = kFactorial(k);

    printf("The factorial of %d is: %lf ", k, factorial); 

    double forward = myEulerForward1(k);

    printf("The Eulers Number: %lf", forward);
}

Answer 1

First i can see an error at sizeof(&n1); Of course n1 is a pointeur. His Value is a RAM address. But for see his Deferenced value, you must use * before n1. & is used for get the adress of something. * is used for get the inside of a pointer.

Use : sizeof(*n1);

In Second, i see that you get a pointer in the prototype of myEulerForward1

double myEulerForward1(int *n1)

It's your compile error. He said that your function need a pointer (an adress) and that you put everything except that.

So when you call this function, you must put a pointer (a RAM adress). And for do that, in the calling of the function, you must use & of course for get the adress of n1 and not his number value.

Use : double forward = myEulerForward1(&k);

Answer 2

Well, yes, it is possible to convert a pointer to an int , and vice versa.

However, you are making a serious mistake in asking that question. There are circumstances where converting an int to a pointer, or vice versa, makes sense. But, in your code, you would be using it as a blunt instrument to force the compiler to accept bad code.

Your compiler is complaining because you have passed an int to a function that expects an int * .

Forcing the issue, by converting that int to a pointer, will stop the compiler complaining, but then you'll (possibly) get some form of runtime error, since the function will receive an invalid pointer.

Your choices are

• remove the * from double myEulerForward1(int *n1) . This will mean, the function expects an int , so your code that passes an int will be correct.
• Call the function as myEulerForward1(&k) which passes an address of k (which is not k converted to a pointer) as a pointer.

Looking at the body of myEulerForward1() there are other problems as well. You need to read up and better understand what sizeof does. Whether your function accepts an int or a pointer ( int * ) the logic of your function is faulty.