I want to know how it will act when the function "add" return a reference to local variable , but it run correctly, no crash

Question

please look at this code firstly . I run it on g++ complier. I want to know how it will act when the function "add" return a reference to local variable , but it run correctly, no crash . Why ?

int & add (int a,int b) {
   int c = a + b ;
   return c ; 
}
int main()
{
   cout<<add(1,2)<<endl;
   int a = add(1,2);
   cout<<a<<endl;
}

It is undefined behavior. Most of the times for smaller programs like this it works but when u take complex programs using heavy memory and recursive etc it is undefined behavior. — Sagar Sakre, Aug 16 '14 at 12:44

score 3 · Answer 1 · answered Aug 16 '14 at 12:39

3

Do not return a reference to a local variable.

It may or may not work.

But will sometime down the line it will bite you.

answered Aug 16 '14 at 12:39

Ed Heal

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score 3 · Accepted Answer · answered Aug 16 '14 at 12:41

3

Because it is undefined behavior. You can't rely on it working and as your code grows it won't.

answered Aug 16 '14 at 12:41

Gerald

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I want to know how it will act when the function "add" return a reference to local variable , but it run correctly, no crash

2 Answers2