decltype question

[bluefrog]

I've copied this snippet of code from a new C++ book, "Learning C++ Functional Programming" by Anggorro.

It doesn't compile though.

#include <iostream>
#include <typeinfo>

const int func1();
const int& func2();

struct X { double d; };

auto main() -> int {
  int i{0};

  const X* x = new X();

  decltype(func1()) f1;
  decltype(func2()) f2;
  decltype(i) i1;

  std::cout << "Declare type func1() " << f1<< "\n";
  std::cout << "Declare type func2() " << f2<< "\n";
  std::cout << "Declare type i " << i1 << std::endl;

  return 0;
}

const int func1() { return 1; }

const int & func2() { return 2; }

The following compile error occurs, using g++ on Linux:

15:21: error: ‘f2’ declared as reference but not initialized

Code sample is from pg 11 of the book.
Can anyone explain why the error?

[goldenchicken]

You can't return a reference to a number, because that would allow you to do:

getRefToNumber() = 6;

Which could result in writing 2 = 6, thus assigning the value of 6 to 2. This isn't possible because 2 is an rvalue and not lvalue.

Your question has nothing to do with decltype.

[goldenchicken]

Second part of the answer: Since your func2() returns a reference, you have to initialize f2. All references should be initialized at creation.

[gunnerfunner]

func2() should return const ref to const int as mentioned above, also the program gives garbage values because f1, f2, il were not initialized:

#include <iostream>
//#include <typeinfo>

const int func1();
const int& func2();

struct X { double d; };

auto main() -> int {
  int i{0};

  const X* x = new X(); //unused variable

  decltype(func1()) f1{};
  decltype(func2()) f2{};
  decltype(i) i1{};

  std::cout << "Declare type func1() " << f1<< "\n";
  std::cout << "Declare type func2() " << f2<< "\n";
  std::cout << "Declare type i " << i1 << std::endl;

  return 0;
}

const int func1() { return 1; }

const int & func2() { const int i {2}; return i; }

for func1() there's the additional warning:

warning: type qualifiers ignored on function return type [-Wignored-qualifiers] 

https://stackoverflow.com/questions/1134237/pedantic-gcc-warning-type-qualifiers-on-function-return-type

[JLBorges]

const int & func2() { const int i {2}; return i; }
This engenders undefined behaviour.

[gunnerfunner]

yeah, you're right i think

[bluefrog]

Hi

thanks for the replies, so as long as the variables to functions that return a reference are initialised it compiles, something like this:

#include <iostream>
#include <typeinfo>

const int func1();
const int& func2();

struct X { double d; };

auto main() -> int {
  int i{0};

  const X* x = new X();

  decltype(func1()) f1;
  decltype(func2()) f2 = f1;
  decltype(i) i1;
  decltype(x->d) d1;
  decltype((x->d)) d2 = d1;

  std::cout << "Declare type func1() " << f1<< "\n";
  std::cout << "Declare type func2() " << f2<< "\n";
  std::cout << "Declare type i " << i1 << std::endl;
  std::cout << "Declare type d1 " << d1 << std::endl;
  std::cout << "Declare type d2 " << d2 << std::endl;

  return 0;
}

const int func1() {
  return 1;
}

const int & func2() {
  int temp{2};
  int *  n = new int;
  *n = temp;
  return *n;
}

I understand f1 & f2, but I don't understand the following 2 declarations:

  decltype(x->d) d1;
  decltype((x->d)) d2 = d1;

which occur on page 12 of the book. Firstly , what is "d", and secondly
why do I have to initialise " decltype((x->d)) d2" to d1 in order for the code to compile ?

[mbozzi]

why do I have to initialise decltype((x->d)) d2

Because the type decltype((x->d)) is a lvalue reference (because the expression x->d is an lvalue, and it is parenthesized.)
http://en.cppreference.com/w/cpp/language/decltype