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c++functionoutputtrace
Sarah_Xx
Sarah_Xx

Reputation: 133

Tracing the output, could not figure it out even when debugging it?

When calling the function f4 how is the function returning 6? i really cant figure out how the function operates shouldnt it just return 1? because of (n-1) 

  #include <iostream>
    #include<cmath>
    #include<fstream>
    using namespace std;
    int x = 3;
    void f1(int, int &);

    int f4(int);

    int main()

    { 
        int x = 5; int y = 10;
        f1(x, y); 
        cout << x << "\t" << y << endl;
        x = 15; y = 20;
        f1(x++, x); 


        cout << x << "\t" << y << endl; 
        x = 3;
        cout << f4(x) << endl; 

        system("pause");
        return 0;
    }

    void f1(int a, int &b)
    {
        a *= 2; b += x;
        cout << a << "\t" << b << endl;
    }

    int f4(int n) {
        if (n == 1 || n == 0) 
            return n;
        else
            return n + f4(n - 1);
    }

Upvotes: 0

Views: 36

Answers (2)

johnathan
johnathan

Reputation: 2355

recursion in a nutshell..

   int f4(int n) {
        if (n == 1 || n == 0) 
            return n;
        else
            return n + f4(n - 1);
 }

Your code states that when n is 1 or 0 just return n, otherwise add n to the result of the function.

that sets up a recursive stack where the first call n = 3 and it recurses. on the next call n = 2 and it recurses. on the next call n = 1 and it returns as well as the rest of the stack leading to 1 + 2 + 3 which is 6.

Upvotes: 2

Bo R
Bo R

Reputation: 2371

The f4 function is recursive. This calling with a number other than 1 or 0 will make it recurse. You call it with 3, so the compiler (simplified) sees

f4(3) => 3 + f4(2) => 3 + 2 + f4(1) => 3 + 2 + 1 => 5 + 1 => 6

Upvotes: 2

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