Recursion with Double Call (C++) -

can please explain me how adds 26?

i confused 'double call'. maybe don't understand recursion think do.

#include <iostream> using namespace std;  int rec(int * n, int max) {     if (max < 0)         return 0;      return n[max] + rec(n, max - 1) + rec(n, max - 2);  }  int main() {    const int max = 5;     int n[] = { 1, 2, 3, 4, 5 };     int f = rec(n, max - 1);      cout << f << endl;     return 0; } 

int f = rec(n, 4)       = n[4] + rec(n, 3) + rec(n, 2)       = 5 + (n[3] + rec(n, 2) + rec(n, 1)) + (n[2] + rec(n, 1) + rec(n, 0))       = 5 + (4 + (n[2] + rec(n, 1) + rec(n, 0)) + (n[1] + rec(n, 0) + rec(n, -1)) + (3 + (n[1] + rec(n, 0) + rec(n, -1)) + (n[0] + rec(n, -1) + rec(n, -2)))       = 5 + (4 + (3 + rec(n, 1) + rec(n, 0)) + (2 + rec(n, 0) + 0) + (3 + (2 + rec(n, 0) + 0) + (1 + 0 + 0))       = 5 + (4 + (3 + (n[1] + rec(n, 0) + rec(n, -1)) + (n[0] + rec(n, -1) + rec(n, -2))) + (2 + (n[0] + rec(n, -1) + rec(n, -2)) + 0) + (3 + (2 + (n[0] + rec(n, -1) + rec(n, -2)) + 0) + 1)       = 5 + (4 + (3 + (2 + (n[0] + rec(n, -1) + rec(n, -2)) + 0) + (1 + 0 + 0)) + (2 + (1 + 0 + 0) + 0) + (3 + (2 + (1 + 0 + 0) + 0) + 1))       = 5 + (4 + (3 + (2 + (1 + 0 + 0) + 0) + 1) + (2 + 1 + 0) + (3 + (2 + 1 + 0) + 1))       = 5 + (4 + (3 + (2 + 1 + 0) + 1) + 3 + (3 + 3 + 1))       = 5 + (4 + (3 + 3 + 1) + 3 + 7)       = 5 + (4 + 7 + 10)       = 5 + 21       = 26 

and suggested above, following code generates entire expression that, when evaluated, equals 26. can played generate step-by-step work.

#include <iostream> #include <string>  using namespace std;  string rec(string * n, int max) {     if (max < 0)         return "0";      auto = rec(n, max - 1), b = rec(n, max - 2);      return "(" + n[max] + " + " + + " + " + b + ")"; }  int main() {     const int max = 5;     string n[] = { "1", "2", "3", "4", "5" };     auto f = rec(n, max - 1);      cout << f << endl;      return 0; }