CODEWITHSUNDEEP
c++
Shravan
Shravan

Reputation: 2918

Explain the program flow in this C++ function

I wrote a strange function to find the factorial of a number

int strange_fact(int n=0)
{
    static int i=n;
    static int j=i;
    if(j>1)     
    {       
        i *= --j;
        strange_fact();         
        return 0x7777;    //<------ This line
    }
    else
        return i;
}

When I commented the 9th line, I was getting the expected output. But I encountered a strange (or maybe not so strange) behaviour after adding that line. What happens when I uncomment it is that program flow reaches line 9 even though a recursive function call precedes it. My question is, how does flow reach line 9?

Upvotes: 2

Views: 248

Answers (4)

Shravan
Shravan

Reputation: 2918

The whole thing, explained by a diagram

Let us consider the case of strange_fact(3)

enter image description here

So, whatever happens in the recursive steps becomes meaningless because ultimately, 0x7777 will be returned when the control comes back to the first call.

Upvotes: 1

galdin
galdin

Reputation: 14074

You need to understand how recursions work.

After calling strange_fact() you've put a return statement. Meaning once the function is executed, it's still going to return 0x7777, which screws the answer up.

Check this out, this should work as expected:

int strange_fact(int n=0)
{
    static int i=n;
    static int j=i;
    if(j>1)     
    {       
        i *= --j;
        strange_fact();         
        // return 0x7777;    // We don't want to return a value yet
    }
    if(j<=1)          // We need to check this condition after strange_fact is executed
        return i;

    // This line will never be executed
    return 0x7777;    //<------ This line
}

If you eliminate the static variables, you'll get something like this:

long fact(long i)
{
    if(i > 0)
        return i * fact(i-1);
    if(i == 0)              // factorial(0) = 1
        return 1;

    throw exception("Negative numbers cannot have factorials"); // Or you can handle it by returning a -ve error code
}

Upvotes: 1

Ed Heal
Ed Heal

Reputation: 60037

This line 

 strange_fact();

does the recursion and throws away the result.

The next next line

return 0x7777;

will return that value finally.

If you remove that line and compiled with the warnings on it will inform you that all paths do not return a value

Upvotes: 2

zoska
zoska

Reputation: 1734

When recursive call to function ends, line 9 will be reached. See this (shorter) example:

int foo(int i) {
    if(i > 0) {
        foo(i-1);
        return 0x7777;
    } else {
      return i;
    }
 }

So when calling foo(1) it will go through first if (because 1 > 0) and foo(0) will be called. Now inside this call (foo(0)) program will go into else barnch (because 0 is not > 0) and foo(0) will return 0. So now we will be back to our first call (foo(1)) and as foo(0) returned, foo(1) will return 0x7777.

Upvotes: 6

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