cout printing in reverse order

Question

Trying overloading operator() in the following example:

#include <iostream>
using namespace std;

class Fib {
  public:
    Fib() : a0_(1), a1_(1) {}
    int operator()();
  private:
    int a0_, a1_;
};
int Fib::operator()() {
    int temp = a0_;
    a0_ = a1_;
    a1_ = temp + a0_;
    return temp;
}

int main() {
    Fib fib;

    cout << fib() <<"," << fib() << "," << fib() << "," << fib() << "," << fib() << "," << fib() << endl;
}

It prints the fib sequence in the reverse order as 8,5,3,2,1,1. I understand the states are kept in () overlading but why the printing is showing up in the reverse order?

score 1 · Answer 1 · answered Apr 08 '15 at 21:24

1

operator << is a some function defined for its parameters. The order of evaluation of function arguments is unspecified. They can be evaluated right to left or left to right. It seems your compiler evaluates them right to left.

answered Apr 08 '15 at 21:24

Vlad from Moscow

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Thanks for the clarification.I was using g++ on cygwin and bsd. – KKG10 Apr 08 '15 at 21:41
@KKG10 You should split this one output statement into several statements. – Vlad from Moscow Apr 09 '15 at 04:54

cout printing in reverse order

1 Answers1