C Returning reference to local variable

Question

Is the following code  func1    correct if it has to return i  I remember reading somewhere that there is a problem when returning reference to a local variable  How is it different from func2      int amp  func1         int i      i   1      return i     int  func2         int  p      p   new int       p   1      return p

User · Answer

A local variable is memory on the stack  that memory is not automatically invalidated when you go out of scope  From a Function deeper nested  higher on the stack in memory   its perfectly safe to access this memory   Once the Function returns and ends though  things get dangerous  Usually the memory is not deleted or overwritten when you return  meaning the memory at that adresss is still containing your data - the pointer seems valid   Until another function builds up the stack and overwrites it  This is why this can work for a while - and then suddenly cease to function after one particularly deeply nested set of functions  or a function with really huge sized or many local objects  reaches that stack-memory again   It even can happen that you reach the same program part again  and overwrite your old local function variable with the new function variable  All this is very dangerous and should be heavily discouraged   Do not use pointers to local objects

User · Answer

A good thing to remember are these simple rules  and they apply to both parameters and return types      Value - makes a copy of the item in question   Pointer - refers to the address of the item in question  Reference - is literally the item in question    There is a time and place for each  so make sure you get to know them  Local variables  as you ve shown here  are just that  limited to the time they are locally alive in the function scope  In your example having a return type of int  and returning  amp i would have been equally incorrect  You would be better off in that case doing this     void func1 int amp  oValue        oValue   1      Doing so would directly change the value of your passed in parameter  Whereas this code     void func1 int oValue        oValue   1      would not  It would just change the value of oValue local to the function call  The reason for this is because you d actually be changing just a  local  copy of oValue  and not oValue itself

User · Answer

This code snippet   int amp  func1         int i      i   1      return i      will not work because you re returning an alias  a reference  to an object with a lifetime limited to the scope of the function call  That means once func1   returns  int i dies  making the reference returned from the function worthless because it now refers to an object that doesn t exist   int main         int amp  p   func1           p is garbage        The second version does work because the variable is allocated on the free store  which is not bound to the lifetime of the function call  However  you are responsible for deleteing the allocated int   int  func2         int  p      p   new int       p   1      return p     int main         int  p   func2           pointee still exists        delete p     get rid of it     Typically you would wrap the pointer in some RAII class and or a factory function so you don t have to delete it yourself   In either case  you can just return the value itself  although I realize the example you provided was probably contrived    int func3         return 1     int main         int v   func3           do whatever you want with the returned value     Note that it s perfectly fine to return big objects the same way func3   returns primitive values because just about every compiler nowadays implements some form of return value optimization   class big object     public      big object    constructor arguments           big object        big object const big object amp  rhs       big object amp  operator  const big object amp  rhs          public methods    private         data members        big object func4         return big object    constructor arguments         int main             no copy is actually made  if your compiler supports RVO     big object o   func4            Interestingly  binding a temporary to a const reference is perfectly legal C     int main            This works  The returned temporary will last as long as the reference exists     const big object amp  o   func4               This does  not  work  It s not legal C   because reference is not const         big object amp  o   func4

[c++] C++ Returning reference to local variable

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