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c++recursionsegmentation-fault
user69514
user69514

Reputation: 27629

Segmentation fault C++ in recursive function

Why do I get a segmentation fault in my recursive function. It happens every time i call it when a value greater than 4 as a parameter

#include <iostream>
#include <limits>

using namespace std;    

int printSeries(int n){
    if(n==1){       
        return 1;
    }
    else if( n==2){     
        return 2;
    }
    else if( n==3){
        return 3;
    }
    else if( n==4){
        return printSeries(1) + printSeries(2) + printSeries(3);
    }
    else{       
        return printSeries(n-3) + printSeries((n-2) + printSeries(n-1));
    }
}


int main(){

        //double infinity = numeric_limits<double>::max();

        for(int i=1; i<=10; i++){
            cout << printSeries(i) << endl;
        }

    return 0;

}

This works fine, but i'm not sure that returns the correct result:

return printSeries(n-3) + printSeries(n-2) + printSeries(n-1);

Upvotes: 3

Views: 3111

Answers (2)

Owen S.
Owen S.

Reputation: 7855

The parentheses problem mentioned above is the source of the infinite recursion. But there's another problem with the code even if you fix the parentheses for case 5 to be like case 4:

printSeries(4) recursively invokes printSeries 3 times.
printSeries(5) recursively invokes printSeries 6 times.
printSeries(6) recursively invokes printSeries 12 times.
printSeries(10) recursively invokes printSeries 156 times.
printSeries(20) recursively invokes printSeries 69747 times.
printSeries(50) recursively invokes printSeries over 6 trillion times.

In other words, your code is doing way more work than it should. Do you see why?

Upvotes: 4

kennytm
kennytm

Reputation: 523724

return printSeries(n-3) + printSeries( (n-2) + printSeries(n-1) );
//                                     ^^^^^^^^^^^^^^^^^^^^^^^^

Incorrect nesting of parenthesis causes infinite recursion, which leads to stack overflow (segfault).

Consider when n = 4,

f(4) = 1 + f(2 + f(3))
     = 1 + f(2 + 3)
     = 1 + f(5)
     = 1 + [ f(2) + f(3 + f(4)) ]
     = ...

Upvotes: 18

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