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Declaration and Initialization with Auto

, 25 Feb 2014 CPOL
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Guidelines about the auto keyword of C++11 for variable declaration


Guidelines about the auto keyword of C++11 for variable declaration.

The guidelines around the usage of the auto keyword in C++11 is controversial. One argument is that “auto hides the type”. In practice, I found that this is not a concern, unless you use “notepad.exe” as your IDE. This post lists a collection of examples and guidelines for local variable declaration and initialization with auto.

Initialization in C++ is a mess.

int i;
float f(3.3);
double d = 4.4;
char c = i + 2;
class A {
  A() : s(), i() {}
  string s;
  int i;
};
string t(); // This is not even a variable declaration!

All the above variables are declared and initialized using different syntax and semantics.
For example, variable i is Default Initialized. In case of int, it is effectively not initialized at all.
Variable f is initialized with 3.3 but it looks more like a function call.
The A::s and A::i variables are Value Initialized, A::i with 0 and A::s with the default constructor of string.
On the other hand, the last one is not a variable definition but a function declaration. The t is a function that takes no arguments and returns a string.
This is anything but consistent. If I was new in C++ land, I would consider this confusing and I may prematurely turn away from learning C++ in favor of some other language. I think it is a pity.

Fortunately with C++11, the language gained a new way of declaring variables. The same code looks like this when using the auto keyword and Braced Initialization:

auto i = 1;
auto f = 3.3f;
auto d = 4.4;
auto c = char{i + 2};
class A {
  A() : s{}, i{} {}
  string s;
  int i;
};

This looks much more consistent and also safer (see the guidelines at the end of this post).

Here is a more elaborate example about the basic usage:

#include <iostream>
#include <string>
#include <typeinfo>
#include <vector>
#include <memory>
using namespace std;

template<typename T>
void print(const T& a) { cout << typeid(T).name() << "=(" << a.i << "," << a.d << "," << a.s << ") " << endl; }

struct A {
  int i;  double d;  string s;
};

struct B {
  explicit B(int ai, string as) : i{ai}, s{as} {}
  virtual ~B() {}
  int i;  double d;  string s;
};

struct C : public B {
  explicit C() : B{1, "C"} {}
};

int main() {                          // Type of declared variable
                                      // ---------------------------
  auto i1 = 11;                       // int
  auto i2 = 12u;                      // unsigned int
  auto i3 = 13.0f;                    // float
  auto i4 = 13.0;                     // double
  auto i5 = "hello";                  // const char[6]
  auto i6 = string{"hello"};          // string
  // auto i7 = int{i3};               // n/a (Error: narrowing from float to int!)

  auto v1 = vector<int>{};            // vector<int> (empty vector)
  auto v2 = vector<int>{1, 2, 3, 4};  // vector<int> (with elements 1, 2, 3, 4)
 
  auto p1 = unique_ptr<B>{new C{}};   // unique_ptr<B>
  auto p2 = shared_ptr<B>{new C{}};   // shared_ptr<B>
  auto p3 = new A{};                  // A*
  delete p3;
  auto p4 = static_cast<B*>(new C{}); // B* (Ugly! Good! Discourages polymorphic raw pointers.)
  delete p4;
 
  // auto a0 = A;                     // n/a (Error: Initialization required! Good!)
  auto a1 = A{};                      // A (Value initialization)
  auto a2 = A{11, 3.3, "A"};          // A (Aggregate initialization)

  // auto b1 = B{};                   // n/a (Error: No such user-defined constructor!)
  auto b2 = B{22, "B"};               // B
  // auto b3 = B{1, 4.4, "B"};        // n/a (Error: No such user-defined constructor!)

  print(a1);  print(a2);
  print(b2);
  return 0;
}

Here is the output of the above program:

1A=(0,0,)
1A=(11,3.3,A)
1B=(22,-0.553252,B)

There are some situations when determining the resulting type of the declaration may look not so obvious, but it all makes sense and remembering the following helps you out:

  • “auto a = expression;” in general:
    • Creates a new object called a.
    • It has the same type as expression.
    • Initialized with the value of expression.
  • “auto a = expression;”
    • a is a new object. The value of expression is copied into a. So, the CV qualifier of the expression is dropped.
  • “auto& a = referencable_expression;” and
  • “auto* a = pointer_expression;” same as
    “auto a = pointer_expression;”

    • This creates a new reference (a reference or a pointer) to the object returned by the expression. Because it is a reference, the CV qualifier of the
      expression is “inherited” to a. For example, this will create a const reference if the expression itself is const. Makes sense!

Here is an example of some “not so obvious” declarations denoting the source type and the resulting type. Pay attention that the first column is the resulting type and the second column is the source type (the “a <- b" notation indicates that we get type a from type b).

#include <iostream>
#include <typeinfo>
using namespace std;

int main() {
  //                                        Result      Source
  //                                    -----------------------------------
  auto i = int{3};                       // int         <-  int

  auto v1 = i;                           // int         <- int
  const auto v2 = i;                     // const int   <- int
  auto& v3 = i;                          // int&        <- int
  const auto& v4 = i;                    // const int&  <- int
  auto v5 = static_cast<const int>(i);   // int         <- const int
  auto v6 = static_cast<const int&>(v3); // int         <- const int&

  auto& t4 = v4;                         // const int&  <- const int& (CV inherited)
  const auto& t5 = v4;                   // const int&  <- const int&
  auto& t6 = v3;                         // int&        <- int&

  auto w1 = v3;                          // int         <- int&
  auto w2 = v4;                          // int         <- const int&

  auto p1 = &i;                          // int*        <- int*
  auto p2 = p1;                          // int*        <- int*
  const auto p3 = p1;                    // int* const  <- int*
  auto& p4 = p1;                         // int*&       <- int*
 
  auto q1 = static_cast<const int*>(&i); // const int*  <- const int* (CV inherited)
  auto q2 = q1;                          // const int*  <- const int* (CV inherited)
  auto q3 = *q1;                         // int         <- const int

  auto* r1 = static_cast<const int*>(&i);// const int*  <- const int* (CV inherited)
  auto* r2 = r1;                         // const int*  <- const int* (CV inherited)
  auto* r3 = *r1;                        // int         <- const int

  auto z1 = {1, 2, 3};                   // std::initializer_list<int>
}

One other important rule to remember is that the Initializer-list Constructor wins over normal constructors. In other words, when in doubt, the compiler picks the Initializer-list Constructor. Here is an example that demonstrates this:

#include <iostream>
#include <string>
#include <typeinfo>
#include <initializer_list>
#include <vector>
using namespace std;

template<typename T>
void print(const T& a) {
  cout << typeid(T).name() << "=(" << a.i << "," << a.d << "," << a.s << ",[";
  for (const auto i : a.v) { cout << i << " "; }
  cout << "])" << endl;
}

struct D {
  explicit D() {}
  explicit D(initializer_list<int> il) : v{il} {}
  int i;  double d;  string s;  vector<int> v;
};

struct C {
  explicit C(int ai, string as) : i{ai}, s{as} {}
  explicit C(int ai1, int ai2) : i{ai1 + ai2} {}
  explicit C(initializer_list<int> il) : v{il} {}
  int i;  double d;  string s;  vector<int> v;
};

int main() {
  auto v1 = vector<int>{};     // Empty vector
  auto v2 = vector<int>{1, 5}; // Vector with values [1, 5]
  auto v3 = vector<int>(1, 5); // Vector with values [1, 1, 1, 1, 1]

  auto d1 = D{};                 // D::D() user-defined
  auto d2 = D{1, 2, 3, 4 ,5, 6}; // D::D(initializer_list<int>)

  auto c1 = C{};                 // C::C(initializer_list<int>), no user-defined C::C()
  // auto c2 = C{33, 4.4, "C"};  // Error:
                                 // Compiler wants C::C(initializer_list<int>)
  auto c3 = C{10, 20};           // C::C(initializer_list<int>)
  auto c4 = C{33, "C"};          // C::C(int, string), compiler cannot deduce init-list
  auto c5 = C{1, 2, 3, 4, 5, 6}; // C::C(initializer_list<int>)

  print(d1);  print(d2);
  print(c1);  print(c3);  print(c4);  print(c5);
  return 0;
}

Here is the output of this example program:

1D=(-1217180992,3.64111e-314,,[])
1D=(-1217352208,-5.02222e-42,,[1 2 3 4 5 6 ])
1C=(-1218706460,-0.318174,,[])
1C=(-1220249883,4.88997e-270,,[10 20 ])
1C=(33,4.85918e-270,C,[])
1C=(-1217346088,4.8697e-270,,[1 2 3 4 5 6 ])

Guidelines

  • Almost Always Auto: Use auto for local variable declaration whenever you can.
    • Safety: Discourages uninitialized variables (Default Initialization).
    • Safety: Discourages using naked pointers (the polymorphic case).
    • Correctness: No automatic narrowing! You get the exact type.
    • Convenience: With auto, the compiler “figures out” the type for you.
    • Convenience: Use standard literal suffixes and prefixes for less typing.
      • Integer suffixes: u (unsigned int), l (long int), ul (unsigned long int).
      • Floating point suffixes: L (long double), f (float).
      • Character prefixes: L’’ (wchar_t), L”” (const wchar_t[]).
  • Almost Always Brace Initializers: Use brace initializers whenever you can.
    • Consistency: Provides one syntax for initialization.
    • Initialization != Function call != Type cast
      • “auto w = Widget(12)”: This looks like a function call, but it’s not. It is a type cast of int to Widget (Explicit Type Conversion).
      • “auto w = Widget{12}”: This looks like a brace initializer and it is a brace initializer.
  • Almost: The compiler will tell you when you cannot use auto.

References

END OF POST 


License

This article, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

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Gabor Fekete

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Comments and Discussions

 
Questionmemory leak PinmemberDusan Paulovic25-Feb-14 7:03 
AnswerRe: memory leak PinmemberGabor Fekete25-Feb-14 22:42 

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