c++ - Why does the following program works fine? I am returning a reference to local variable from a function -

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• can local variable's memory accessed outside scope? 20 answers

i baffled how come following program works fine. returning reference local variable function , reference assigned value more once. expect compiler throw error reference assignment.

#include <iostream> using namespace std;  int& getnum() {     int mynum = 89;     return mynum; }  int& getanothernum() {     int mynum = 1000;     return mynum; }  int main() {     int& value1 = getanothernum();        cout << "value1 value is: " << value1 << endl;      value1 = getnum();          cout << "value1 value is: " << value1 << endl;  return 0; } 

that undefined behavior. per §1.3.24, undefined behavior described as:

behavior international standard imposes no requirements

contrary popular belief doesn't mean produce error. standard imposes no requirements whatsoever.

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why did allow "value1 = getnum(); ". value1 reference, assigned something.

because in:

value1 = getnum(); 

you not reassigning reference. calling operator= on int copied value of returned value of getnum value1.