CODEWITHSUNDEEP
c++c++11
user2836797
user2836797

Reputation:

std::map::at returns warning "returning reference to temporary" somewhere in the call stack

I am looking up the value of a key from a std::map. I get the following error inside Dataset::getProperty():

returning reference to temporary [-Wreturn-local-addr]

Where in the call stack is the temporary created? I thought std::map::at returned a reference to an lvalue, and because I continously return reference-to-const that the original caller would have a reference to the lvalue.

Dataset

class Dataset
{
  private:
    PropertyCollection properties;

  public:
    const std::string & getProperty(const std::string & key) const;

  ...
}

const std::string & Dataset::getProperty(const std::string & key) const
{
  // WARNING: returning reference to temporary [-Wreturn-local-addr]
  return properties.getProperty(key);
}

PropertyCollection

class PropertyCollection
{
  private:
    std::map<std::string, std::string> properties;

  public:
    const bmd2::string & getProperty(const bmd2::string & key) const

  ...
}

const std::string & PropertyCollection::getProperty(const std::string & key) 
      const
{
  try {
    return properties.at(key);
  } catch (std::out_of_range & e) {
    ...
  }

Main

int main() {
  ...
  std::string key ("Cats");
  std::string value = dataset.getProperty(key);
}

Upvotes: 2

Views: 661

Answers (1)

Sebastian Hoffmann
Sebastian Hoffmann

Reputation: 11502

getProperty returns a bmd2::string not a std::string. Thus a std::string must be implicitely constructed/converted from a bmd2::string. This std::string is then of course a new temporary object.

While its actually not illegal to return a reference to a temporary, using such a reference (if out of scope) will result in UB, thus the compiler warns you.

Upvotes: 10

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