我在C代码中遇到错误：二进制表达式的操作数无效（'long long（int）'和'int'）我不知道如何修复

Question

I am getting the above stated error in my C code. I am also getting a warning about the same thing saying the pointer to integer conversion is incompatible. The objective of the assignment is to print the factorial and two approximating factorials of the integer the user inputs. Could I just get help on how to fix the errors?

Here's the code:

#include <stdio.h>
#include <math.h>

#define PI 3.141592653589793
#define e 2.71828
#define TRUE 1
#define FALSE 0

long long fact(int n);
double stirling1(int n);
double stirling2(int n);

/*--------------------------- main function -------------------------------
Purpose: The purpose of this program is to give the user the value of the 
factorial of the nonnegative integer the user inputs.
---------------------------------------------------------------------------*/

int main ()
{
char ch = 'y';
int flag, again;
int n;

again = TRUE;

while ( again == TRUE )
{
    printf("Please enter a nonnegative integer:"); //ask the user for the input
    flag = scanf("%d\n\n", &n);

    if ( flag != 1 ) //run this statement if the user inputs a noninteger
    {
        printf("Input must be an integer.");
        continue;
    }

    else if ( n < 0 ) //run this statement if the user inputs a negative integer
    {
        printf("The factorial is undefined...");
        continue;
    }

    else if ( n <= 14 ) //run this statement if the user inputs an integer less than or equal to 14
    {
        printf("Number     Factorial    Aprroximation          Aproxximation2\n------------------------------------------------------------------\n"); //prints the header for first table
        for ( n = 0; n <= 14; n++ )
            {
                ++n;
                printf("%d%1411lld%9e%9e\n", n, fact( n ), stirling1( n ), stirling2( n )); //calls functions to input factorials
            }
    }

    else //run this statement if the user inputs a number greater than 14
    {
        printf("Number   Approximation1         Approximation2\n-----------------------------------------------------\n"); //prints the header for second table
        for ( n = 0; n > 14; n++ )
        {
            ++n;
            printf("%3d%9e%e\n", n, stirling1( n ), stirling2( n )); //calls functions to input approximate factorials
        }
    }

    printf("Do you want to compute another factorial? (y/n):"); //ask user if they want another factorial of a different number
    scanf("%c\n", &ch);

    if (ch != 'y') 
        again = FALSE; //if user does not input 'y' then do not compute another factorial
}

printf( "**** Program Terminated ****" ); //ends program

}

long long fact( int n ) //function to find exact factorial
{
    fact *= n; //equation for exact factorial

    return fact; //return exact factorial to main
}

double stirling1( int n ) //function to find first approximate factorial
{
    n = pow( n, n ) * pow( e, -n ) * sqrt( 2 * PI * n); //equation to find    first approximate factorial

    return n; //return approximate factorial to main
}

double stirling2( int n ) //function to find second approximate factorial
{
    n = pow( n, n ) * pow( e, -n ) * sqrt( 2 * PI * n) * ( 1 + ( 1 / (12 * n) ));
    //equation to find second approximate factorial

    return n; //return approximate factorial to main
}

Here's the error:

Lab_Assignment_4_Sarah_H.c:97:7: error: invalid operands to binary
  expression ('long long (int)' and 'int')
    fact *= n; //equation for exact factorial
    ~~~~ ^  ~
Lab_Assignment_4_Sarah_H.c:99:9: warning: incompatible pointer to integer
  conversion returning 'long long (int)' from a function with result type
  'long long' [-Wint-conversion]
    return fact; //return exact factorial to main
           ^~~~

Answer 1

You're attempting to use the name of a function as a variable, using it in a context where an integer type is expected.

What you want to do is call the function again with a value of n-1 , multiply the result by n , then return the result of the multiplication.

return n * fact(n-1);

There's still one piece missing. If you just do that, there's nothing to stop the recursive function calls. The function will keep calling itself until you run out of stack space.

What's needed is a base case which stops the recursion.

if (n <= 1) {
    return 1;
} else {
    return n * fact(n-1);
}