Cannot do a simple function using call by value method in C - c

Question

I am trying to test this method (call by value) i recently learned, but for unkown reasons for me, the code isn't working, can someone explain me why? (sorry for the mess by the way)
#include <stdio.h>
#include <stdlib.h>
void fact(int i);
int main() {
int x;
scanf("%d", x);
//function called:
fact(x);
printf("value of x! = %d", x);
return 0;
}
//factorial definition:
void fact(int i){
int j=0;
for(i=1; j>1; j--){
i= i * j;
return;
}
}

Answer 1

If you are using a call to a function by value to make calculations with the passed argument then to get the result of the calculations in the caller the function must return a value. That is its return type should not be void.
The definition of the function fact does not make sense. In the for loop the parameter i is reassigned to 1. The loop itself never iterates because another variable j is set to 0. So it can not be greater than 1.
int j=0;
for(i=1; j>1; j--){
The function can be defined the following way
//factorial definition:
unsigned long long int fact( unsigned int i )
{
unsigned long long int f = 1;
while ( i > 1 ) f *= i--;
return f;
}
The factorial is defined for unsigned natural integers.
In main you need to declare a variable of the type unsigned int like
unsigned int x;
This call of scanf
scanf("%d", x);
is incorrect. The function needs to accept its integer argument by reference to change it.
scanf( "%u", &x );
To get the result of the function fact you have to introduce one more variable
unsigned long long int result = fact( x );
And to output it you should write
printf( "value of %u! = %llu\n", x, result );
Here is a demonstrative program.
#include <stdio.h>
unsigned long long int fact( unsigned int );
int main(void)
{
unsigned int x;
scanf( "%u", &x );
unsigned long long int result = fact( x );
printf( "value of %u! = %llu\n", x, result );
return 0;
}
//factorial definition:
unsigned long long int fact( unsigned int i )
{
unsigned long long int f = 1;
while ( i > 1 ) f *= i--;
return f;
}
If to enter the number 20 then the function output will be
value of 20! = 2432902008176640000
Pay attention to that 20 is the maximum number the factorial of which can be stored in an object of the type unsigned long long int.
If you want that the function accepted its argument by reference and returned the result in the argument then you have to declare the variable x as having the type unsigned long long int.
Here is a demonstrative program.
#include <stdio.h>
void fact( unsigned long long int * );
int main(void)
{
unsigned long long int x;
scanf( "%llu", &x );
printf( "value of %llu! = ", x );
fact( &x );
printf( "%llu\n", x );
return 0;
}
//factorial definition:
void fact( unsigned long long int *x )
{
unsigned long long int f = 1;
while ( *x > 1 ) f *= ( *x )--;
*x = f;
}
If to enter 20 then the program output will be the same as shown above.
value of 20! = 2432902008176640000

Answer 2

Scanf takes a pointer to the variable, to return something from a void function you must call it by reference with a pointer to the return variable and not by its value.

Answer 3

Problem 1:
scanf("%d", x);
Should be:
scanf("%d", &x);
The & is very important.
Problem 2
In function fact, you start j at 0:
int j=0;
Your loop only runs while j>1:
for(i=1; j>1; j--){
Tell me: When is 0 ever >1??
Your loop will never run.
Problem 3:
You expect that calling fact(x); will change the value of x.
In C, a function cannot change one of its arguments; arguments are passed "by-copy".
If you wish to change an object (variable) in the caller, you must pass a pointer to it:
void fact(int* i);
[...]
//function called:
fact(&x);
[...]
//factorial definition:
void fact(int* i){
int j=0;
for(*i=1; j>1; j--){
*i= *i * j;
return;
}
}