Reputation: 11
#include <stdio.h>
int main (void)
{
int n, x;
int factorial (int n)
{
if (x<=0)
{
printf("x equals: ");
return 1;
}
else
{
return n * factorial (n-1);
}
f(x)=f(x-1)+2; //says error is here
}
return 0;
}
I've tried some things and can't get it to work. I could just be overtired and looking past the smallest thing but help would be much appreciated! Thanks :)
Upvotes: 1
Views: 221
Reputation: 32530
You cannot declare a function definition inside of main()
or any other function ... function definitions have to be stand-alone and cannot have embedded function definitions inside of them.
Also I'm not sure what you're doing on the line that you've marked as an error since f()
is not a defined function, so you can't call it. Furthermore, it would need to return some type of l-value, such as a pointer to a static variable declared inside the function, or a pointer passed by reference to the function and even then the syntax is not right since there would be a required dereference ... so basically you can't do what you're doing on that line.
To get something that compiles, try
#include <stdio.h>
int factorial (int n)
{
if (n <= 0)
{
return 1;
}
else
{
return n * factorial (n-1);
}
}
int main (void)
{
int x;
x = factorial(5);
printf("Factorial of 5 is equal to %d", x);
return 0;
}
Upvotes: 6
Reputation: 92874
A function cannot be defined inside another function. However gcc allows it as an extension. You have defined a function named factorial
but are trying to use f
which hasn't been declared anywhere.
Upvotes: 1
Reputation: 6602
Use indentation to see possible problems with scope:
#include <stdio.h>
int main (void)
{
int n, x;
int factorial (int n)
{
if (x<=0)
{
printf("x equals: ");
return 1;
}
else
{
return n * factorial (n-1);
}
f(x)=f(x-1)+2; //says error is here
}
return 0;
}
As far as I can remember, C doesn't have closures.
Upvotes: 1