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What happens when one uses auto with two declarations in C++11?
(1 answer)
Closed 10 months ago.
I know I can write
if (int a = 1; /* whatever */) {}
and even
if (int a = 1, b{3}; /* whatever */) {}
but how can I declare, say, a of type int and b of type std::string?
Such a thing doesn't work:
if (auto a = 1, b{"ciaos"s}; /* whatever */) {}
I've not included a standard, because I'm interested in the answer in general, even though realistically I'd make use of the answer in the context of c++17.
And, if such a thing is not possible, is there any precise reason why (hence language-lawyer)?
You are only allowed one variable declaration statement in a if statement and each variable declaration statement can only declare a single type. This is convered in [stmt.if]/3 where is shows the grammar for the if statement you are trying to use is
if constexpr(opt) ( init-statement condition ) statement
and init-statement can be a simple-declaration and that contains a init-declarator-list which only allows a single declarator. Normally this means just a single type, but pointers (*) and references (&) get applied to the variable name, not the type name so you can have T, T*, and/or T& variables declared in a single init-declarator-list i.e., int a = 42, *b = &a, &c = a;
As a workaround, you can leverage structured bindings and CTAD (to reduce verbosity) in conjunction with std::tuple to get a syntax like
int main()
{
using namespace std::string_literals;
if (auto [a, b] = std::tuple{42, "string"s}; a)
{
std::cout << b;
}
}
which outputs
string
This is something that has always been bugging me as a feature of C++ lambda expressions: The type of a C++ lambda expression is unique and anonymous, I simply cannot write it down. Even if I create two lambdas that are syntactically exactly the same, the resulting types are defined to be distinct. The consequence is, that a) lambdas can only be passed to template functions that allow the compile time, unspeakable type to be passed along with the object, and b) that lambdas are only useful once they are type erased via std::function<>.
Ok, but that's just the way C++ does it, I was ready to write it off as just an irksome feature of that language. However, I just learned that Rust seemingly does the same: Each Rust function or lambda has a unique, anonymous type. And now I'm wondering: Why?
So, my question is this:
What is the advantage, from a language designer point of view, to introduce the concept of a unique, anonymous type into a language?
Many standards (especially C++) take the approach of minimizing how much they demand from compilers. Frankly, they demand enough already! If they don't have to specify something to make it work, they have a tendency to leave it implementation defined.
Were lambdas to not be anonymous, we would have to define them. This would have to say a great deal about how variables are captured. Consider the case of a lambda [=](){...}. The type would have to specify which types actually got captured by the lambda, which could be non-trivial to determine. Also, what if the compiler successfully optimizes out a variable? Consider:
static const int i = 5;
auto f = [i]() { return i; }
An optimizing compiler could easily recognize that the only possible value of i that could be captured is 5, and replace this with auto f = []() { return 5; }. However, if the type is not anonymous, this could change the type or force the compiler to optimize less, storing i even though it didn't actually need it. This is a whole bag of complexity and nuance that simply isn't needed for what lambdas were intended to do.
And, on the off-case that you actually do need a non-anonymous type, you can always construct the closure class yourself, and work with a functor rather than a lambda function. Thus, they can make lambdas handle the 99% case, and leave you to code your own solution in the 1%.
Deduplicator pointed out in comments that I did not address uniqueness as much as anonymity. I am less certain of the benefits of uniqueness, but it is worth noting that the behavior of the following is clear if the types are unique (action will be instantiated twice).
int counter()
{
static int count = 0;
return count++;
}
template <typename FuncT>
void action(const FuncT& func)
{
static int ct = counter();
func(ct);
}
...
for (int i = 0; i < 5; i++)
action([](int j) { std::cout << j << std::endl; });
for (int i = 0; i < 5; i++)
action([](int j) { std::cout << j << std::endl; });
If the types were not unique, we would have to specify what behavior should happen in this case. That could be tricky. Some of the issues that were raised on the topic of anonymity also raise their ugly head in this case for uniqueness.
Lambdas are not just functions, they are a function and a state. Therefore both C++ and Rust implement them as an object with a call operator (operator() in C++, the 3 Fn* traits in Rust).
Basically, [a] { return a + 1; } in C++ desugars to something like
struct __SomeName {
int a;
int operator()() {
return a + 1;
}
};
then using an instance of __SomeName where the lambda is used.
While in Rust, || a + 1 in Rust will desugar to something like
{
struct __SomeName {
a: i32,
}
impl FnOnce<()> for __SomeName {
type Output = i32;
extern "rust-call" fn call_once(self, args: ()) -> Self::Output {
self.a + 1
}
}
// And FnMut and Fn when necessary
__SomeName { a }
}
This means that most lambdas must have different types.
Now, there are a few ways we could do that:
With anonymous types, which is what both languages implement. Another consequence of that is that all lambdas must have a different type. But for language designers, this has a clear advantage: Lambdas can be simply described using other already existing simpler parts of the language. They are just syntax sugar around already existing bits of the language.
With some special syntax for naming lambda types: This is however not necessary since lambdas can already be used with templates in C++ or with generics and the Fn* traits in Rust. Neither language ever force you to type-erase lambdas to use them (with std::function in C++ or Box<Fn*> in Rust).
Also note that both languages do agree that trivial lambdas that do not capture context can be converted to function pointers.
Describing complex features of a languages using simpler feature is pretty common. For example both C++ and Rust have range-for loops, and they both describe them as syntax sugar for other features.
C++ defines
for (auto&& [first,second] : mymap) {
// use first and second
}
as being equivalent to
{
init-statement
auto && __range = range_expression ;
auto __begin = begin_expr ;
auto __end = end_expr ;
for ( ; __begin != __end; ++__begin) {
range_declaration = *__begin;
loop_statement
}
}
and Rust defines
for <pat> in <head> { <body> }
as being equivalent to
let result = match ::std::iter::IntoIterator::into_iter(<head>) {
mut iter => {
loop {
let <pat> = match ::std::iter::Iterator::next(&mut iter) {
::std::option::Option::Some(val) => val,
::std::option::Option::None => break
};
SemiExpr(<body>);
}
}
};
which while they seem more complicated for a human, are both simpler for a language designer or a compiler.
Cort Ammon's accepted answer is good, but I think there's one more important point to make about implementability.
Suppose I have two different translation units, "one.cpp" and "two.cpp".
// one.cpp
struct A { int operator()(int x) const { return x+1; } };
auto b = [](int x) { return x+1; };
using A1 = A;
using B1 = decltype(b);
extern void foo(A1);
extern void foo(B1);
The two overloads of foo use the same identifier (foo) but have different mangled names. (In the Itanium ABI used on POSIX-ish systems, the mangled names are _Z3foo1A and, in this particular case, _Z3fooN1bMUliE_E.)
// two.cpp
struct A { int operator()(int x) const { return x + 1; } };
auto b = [](int x) { return x + 1; };
using A2 = A;
using B2 = decltype(b);
void foo(A2) {}
void foo(B2) {}
The C++ compiler must ensure that the mangled name of void foo(A1) in "two.cpp" is the same as the mangled name of extern void foo(A2) in "one.cpp", so that we can link the two object files together. This is the physical meaning of two types being "the same type": it's essentially about ABI-compatibility between separately compiled object files.
The C++ compiler is not required to ensure that B1 and B2 are "the same type." (In fact, it's required to ensure that they're different types; but that's not as important right now.)
What physical mechanism does the compiler use to ensure that A1 and A2 are "the same type"?
It simply burrows through typedefs, and then looks at the fully qualified name of the type. It's a class type named A. (Well, ::A, since it's in the global namespace.) So it's the same type in both cases. That's easy to understand. More importantly, it's easy to implement. To see if two class types are the same type, you take their names and do a strcmp. To mangle a class type into a function's mangled name, you write the number of characters in its name, followed by those characters.
So, named types are easy to mangle.
What physical mechanism might the compiler use to ensure that B1 and B2 are "the same type," in a hypothetical world where C++ required them to be the same type?
Well, it couldn't use the name of the type, because the type doesn't have a name.
Maybe it could somehow encode the text of the body of the lambda. But that would be kind of awkward, because actually the b in "one.cpp" is subtly different from the b in "two.cpp": "one.cpp" has x+1 and "two.cpp" has x + 1. So we'd have to come up with a rule that says either that this whitespace difference doesn't matter, or that it does (making them different types after all), or that maybe it does (maybe the program's validity is implementation-defined, or maybe it's "ill-formed no diagnostic required"). Anyway, mangling lambda types the same way across multiple translation units is certainly a harder problem than mangling named types like A.
The easiest way out of the difficulty is simply to say that each lambda expression produces values of a unique type. Then two lambda types defined in different translation units definitely are not the same type. Within a single translation unit, we can "name" lambda types by just counting from the beginning of the source code:
auto a = [](){}; // a has type $_0
auto b = [](){}; // b has type $_1
auto f(int x) {
return [x](int y) { return x+y; }; // f(1) and f(2) both have type $_2
}
auto g(float x) {
return [x](int y) { return x+y; }; // g(1) and g(2) both have type $_3
}
Of course these names have meaning only within this translation unit. This TU's $_0 is always a different type from some other TU's $_0, even though this TU's struct A is always the same type as some other TU's struct A.
By the way, notice that our "encode the text of the lambda" idea had another subtle problem: lambdas $_2 and $_3 consist of exactly the same text, but they should clearly not be considered the same type!
By the way, C++ does require the compiler to know how to mangle the text of an arbitrary C++ expression, as in
template<class T> void foo(decltype(T())) {}
template void foo<int>(int); // _Z3fooIiEvDTcvT__EE, not _Z3fooIiEvT_
But C++ doesn't (yet) require the compiler to know how to mangle an arbitrary C++ statement. decltype([](){ ...arbitrary statements... }) is still ill-formed even in C++20.
Also notice that it's easy to give a local alias to an unnamed type using typedef/using. I have a feeling that your question might have arisen from trying to do something that could be solved like this.
auto f(int x) {
return [x](int y) { return x+y; };
}
// Give the type an alias, so I can refer to it within this translation unit
using AdderLambda = decltype(f(0));
int of_one(AdderLambda g) { return g(1); }
int main() {
auto f1 = f(1);
assert(of_one(f1) == 2);
auto f42 = f(42);
assert(of_one(f42) == 43);
}
EDITED TO ADD: From reading some of your comments on other answers, it sounds like you're wondering why
int add1(int x) { return x + 1; }
int add2(int x) { return x + 2; }
static_assert(std::is_same_v<decltype(add1), decltype(add2)>);
auto add3 = [](int x) { return x + 3; };
auto add4 = [](int x) { return x + 4; };
static_assert(not std::is_same_v<decltype(add3), decltype(add4)>);
That's because captureless lambdas are default-constructible. (In C++ only as of C++20, but it's always been conceptually true.)
template<class T>
int default_construct_and_call(int x) {
T t;
return t(x);
}
assert(default_construct_and_call<decltype(add3)>(42) == 45);
assert(default_construct_and_call<decltype(add4)>(42) == 46);
If you tried default_construct_and_call<decltype(&add1)>, t would be a default-initialized function pointer and you'd probably segfault. That's, like, not useful.
(Adding to Caleth's answer, but too long to fit in a comment.)
The lambda expression is just syntactic sugar for an anonymous struct (a Voldemort type, because you can't say its name).
You can see the similarity between an anonymous struct and the anonymity of a lambda in this code snippet:
#include <iostream>
#include <typeinfo>
using std::cout;
int main() {
struct { int x; } foo{5};
struct { int x; } bar{6};
cout << foo.x << " " << bar.x << "\n";
cout << typeid(foo).name() << "\n";
cout << typeid(bar).name() << "\n";
auto baz = [x = 7]() mutable -> int& { return x; };
auto quux = [x = 8]() mutable -> int& { return x; };
cout << baz() << " " << quux() << "\n";
cout << typeid(baz).name() << "\n";
cout << typeid(quux).name() << "\n";
}
If that is still unsatisfying for a lambda, it should be likewise unsatisfying for an anonymous struct.
Some languages allow for a kind of duck typing that is a little more flexible, and even though C++ has templates that doesn't really help in making a object from a template that has a member field that can replace a lambda directly rather than using a std::function wrapper.
Why design a language with unique anonymous types?
Because there are cases where names are irrelevant and not useful or even counter-productive. In this case the ability abstract out their existence is useful because it reduces name pollution, and solves one of the two hard problems in computers science (how to name things). For the same reason, temporary objects are useful.
lambda
The uniqueness is not a special lambda thing, or even special thing to anonymous types. It applies to named types in the language as well. Consider following:
struct A {
void operator()(){};
};
struct B {
void operator()(){};
};
void foo(A);
Note that I cannot pass B into foo, even though the classes are identical. This same property applies to unnamed types.
lambdas can only be passed to template functions that allow the compile time, unspeakable type to be passed along with the object ... erased via std::function<>.
There's a third option for a subset of lambdas: Non-capturing lambdas can be converted to function pointers.
Note that if the limitations of an anonymous type are a problem for a use case, then the solution is simple: A named type can be used instead. Lambdas don't do anything that cannot be done with a named class.
C++ lambdas need distinct types for distinct operations, as C++ binds statically. They are only copy/move-constructable, so mostly you don't need to name their type. But that's all somewhat of an implementation detail.
I'm not sure if C# lambdas have a type, as they are "anonymous function expressions", and they immediately get converted to a compatible delegate type or expression tree type. If the do, it's probably an unpronouncable type.
C++ also has anonymous structs, where each definition leads to a unique type. Here the name isn't unpronouncable, it simply doesn't exist as far as the standard is concerned.
C# has anonymous data types, which it carefully forbids from escaping from the scope they are defined. The implementation gives a unique, unpronouncable name to those too.
Having an anonymous type signals to the programmer that they shouldn't poke around inside their implementation.
Aside:
You can give a name to a lambda's type.
auto foo = []{};
using Foo_t = decltype(foo);
If you don't have any captures, you can use a function pointer type
void (*pfoo)() = foo;
Why use anonymous types?
For types that are automatically generated by the compiler, the choice is to either (1) honor a user's request for the name of the type, or (2) let the compiler choose one on its own.
In the former case, the user is expected to explicitly provide a name each time such a construct appears (C++/Rust: whenever a lambda is defined; Rust: whenever a function is defined). This is a tedious detail for the user to provide each time, and in the majority of cases the name is never referred to again. Thus it make sense to let the compiler figure out a name for it automatically, and use existing features such as decltype or type inference to reference the type in the few places where it is needed.
In the latter case, the compiler need to choose a unique name for the type, which would probably be an obscure, unreadable name such as __namespace1_module1_func1_AnonymousFunction042. The language designer could specify precisely how this name is constructed in glorious and delicate detail, but this needlessly exposes an implementation detail to the user that no sensible user could rely upon, since the name is no doubt brittle in the face of even minor refactors. This also unnecessarily constrains the evolution of the language: future feature additions may cause the existing name generation algorithm to change, leading to backward compatibility issues. Thus, it makes sense to simply omit this detail, and assert that the auto-generated type is unutterable by the user.
Why use unique (distinct) types?
If a value has a unique type, then an optimizing compiler can track a unique type across all its use sites with guaranteed fidelity. As a corollary, the user can then be certain of the places where the provenance of this particular value is full known to the compiler.
As an example, the moment the compiler sees:
let f: __UniqueFunc042 = || { ... }; // definition of __UniqueFunc042 (assume it has a nontrivial closure)
/* ... intervening code */
let g: __UniqueFunc042 = /* some expression */;
g();
the compiler has full confidence that g must necessarily originate from f, without even knowing the provenance of g. This would allow the call to g to be devirtualized. The user would know this too, since the user has taken great care to preserve the unique type of f through the flow of data that led to g.
Necessarily, this constrains what the user can do with f. The user is not at liberty to write:
let q = if some_condition { f } else { || {} }; // ERROR: type mismatch
as that would lead to the (illegal) unification of two distinct types.
To work around this, the user could upcast the __UniqueFunc042 to the non-unique type &dyn Fn(),
let f2 = &f as &dyn Fn(); // upcast
let q2 = if some_condition { f2 } else { &|| {} }; // OK
The trade-off made by this type erasure is that uses of &dyn Fn() complicate the reasoning for the compiler. Given:
let g2: &dyn Fn() = /*expression */;
the compiler has to painstakingly examine the /*expression */ to determine whether g2 originates from f or some other function(s), and the conditions under which that provenance holds. In many circumstances, the compiler may give up: perhaps human could tell that g2 really comes from f in all situations but the path from f to g2 was too convoluted for the compiler to decipher, resulting in a virtual call to g2 with pessimistic performance.
This becomes more evident when such objects delivered to generic (template) functions:
fn h<F: Fn()>(f: F);
If one calls h(f) where f: __UniqueFunc042, then h is specialized to a unique instance:
h::<__UniqueFunc042>(f);
This enables the compiler to generate specialized code for h, tailored for the particular argument of f, and the dispatch to f is quite likely to be static, if not inlined.
In the opposite scenario, where one calls h(f) with f2: &Fn(), the h is instantiated as
h::<&Fn()>(f);
which is shared among all functions of type &Fn(). From within h, the compiler knows very little about an opaque function of type &Fn() and so could only conservatively call f with a virtual dispatch. To dispatch statically, the compiler would have to inline the call to h::<&Fn()>(f) at its call site, which is not guaranteed if h is too complex.
First, lambda without capture are convertible to a function pointer. So they provide some form of genericity.
Now why lambdas with capture are not convertible to pointer? Because the function must access the state of the lambda, so this state would need to appear as a function argument.
To avoid name collisions with user code.
Even two lambdas with same implementation will have different types. Which is okay because I can have different types for objects too even if their memory layout is equal.
I haven't been able to find a good explanation of decltype. Please tell me, as a beginning programmer, what it does and why it is useful.
For example, I am reading a book that asked the following question. Can someone explain to me the answer and why, along with some good (beginner-level) examples?
What would be the type of each variable and what value would each variable have when the code finishes?
int a = 3, b = 4;
decltype(a) c = a;
decltype((b)) d = a;
++c;
++d;
A line-by-line explanation would be very helpful.
decltype is a way to specify a type: You give it an expression, and decltype gives you back a type which corresponds to the type of the expression. Specifically, decltype(e) is the following type:
If e is the name of a variable, i.e. an "id-expression", then the resulting type is the type of the variable.
Otherwise, if e evaluates to an lvalue of type T, then the resulting type is T &, and if e evaluates to an rvalue of type T, then the resulting type is T.
Combining these rules with reference collapsing rules allows you to make sense of decltype(e) &&, which is always a "suitable" reference. (C++14 also adds decltype(auto) to give you the type-deduction of auto combined with the value category semantics of decltype.)
Examples:
int foo();
int n = 10;
decltype(n) a = 20; // a is an "int" [id-expression]
decltype((n)) b = a; // b is an "int &" [(n) is an lvalue]
decltype(foo()) c = foo(); // c is an "int" [rvalue]
decltype(foo()) && r1 = foo(); // int &&
decltype((n)) && r2 = n; // int & [& && collapses to &]
It might be worth stressing the difference between auto and decltype: auto works on types, and decltype works on expressions.
You shouldn't be seeing or using decltype in "day-to-day" programming. It is most useful in generic (templated) library code, where the expression in question is not known and depends on a paramater. (By contrast, auto may be used generously all over the place.) In short, if you're new to programming, you probably won't need to use decltype for some time.
From all the material I used to learn C++, auto has always been a weird storage duration specifier that didn't serve any purpose. But just recently, I encountered code that used it as a type name in and of itself. Out of curiosity I tried it, and it assumes the type of whatever I happen to assign to it!
Suddenly STL iterators and, well, anything at all that uses templates is 10 fold easier to write. It feels like I'm using a 'fun' language like Python.
Where has this keyword been my whole life? Will you dash my dreams by saying it's exclusive to visual studio or not portable?
auto was a keyword that C++ "inherited" from C that had been there nearly forever, but virtually never used because there were only two possible conditions: either it wasn't allowed, or else it was assumed by default.
The use of auto to mean a deduced type was new with C++11.
At the same time, auto x = initializer deduces the type of x from the type of initializer the same way as template type deduction works for function templates. Consider a function template like this:
template<class T>
int whatever(T t) {
// point A
};
At point A, a type has been assigned to T based on the value passed for the parameter to whatever. When you do auto x = initializer;, the same type deduction is used to determine the type for x from the type of initializer that's used to initialize it.
This means that most of the type deduction mechanics a compiler needs to implement auto were already present and used for templates on any compiler that even sort of attempted to implement C++98/03. As such, adding support for auto was apparently fairly easy for essentially all the compiler teams--it was added quite quickly, and there seem to have been few bugs related to it either.
When this answer was originally written (in 2011, before the ink was dry on the C++ 11 standard) auto was already quite portable. Nowadays, it's thoroughly portable among all the mainstream compilers. The only obvious reasons to avoid it would be if you need to write code that's compatible with a C compiler, or you have a specific need to target some niche compiler that you know doesn't support it (e.g., a few people still write code for MS-DOS using compilers from Borland, Watcom, etc., that haven't seen significant upgrades in decades). If you're using a reasonably current version of any of the mainstream compilers, there's no reason to avoid it at all though.
More recent revisions of the standard have added a few new places that auto can be used. Starting with C++14, you can use auto for the type of a parameter to a lambda:
[](auto s) { return s + 1; }
This does essentially the same thing as the example above--even though it doesn't explicitly use template syntax, this is basically a template that deduces the type of the parameter, and instantiates the template over that type.
That was convenient and useful enough that in C++20, the same capability was added for normal functions, not just lambdas.
But, just as before all of this really comes down to using the same basic type deduction mechanism as we've had for function templates since C++98. auto allows that to be used in more places, and more conveniently, but the underlying heavy lifting remains the same.
It's just taking a generally useless keyword and giving it a new, better functionality. It's standard in C++11, and most C++ compilers with even some C++11 support will support it.
For variables, specifies that the type of the variable that is being declared will be automatically deduced from its initializer. For functions, specifies that the return type is a trailing return type or will be deduced from its return statements (since C++14).
Syntax
auto variable initializer (1) (since C++11)
auto function -> return type (2) (since C++11)
auto function (3) (since C++14)
decltype(auto) variable initializer (4) (since C++14)
decltype(auto) function (5) (since C++14)
auto :: (6) (concepts TS)
cv(optional) auto ref(optional) parameter (7) (since C++14)
Explanation
When declaring variables in block scope, in namespace scope, in initialization statements of for loops, etc., the keyword auto may be used as the type specifier.
Once the type of the initializer has been determined, the compiler determines the type that will replace the keyword auto using the rules for template argument deduction from a function call (see template argument deduction#Other contexts for details). The keyword auto may be accompanied by modifiers, such as const or &, which will participate in the type deduction. For example, given const auto& i = expr;, the type of i is exactly the type of the argument u in an imaginary template template<class U> void f(const U& u) if the function call f(expr) was compiled. Therefore, auto&& may be deduced either as an lvalue reference or rvalue reference according to the initializer, which is used in range-based for loop.
If auto is used to declare multiple variables, the deduced types must match. For example, the declaration auto i = 0, d = 0.0; is ill-formed, while the declaration auto i = 0, *p = &i; is well-formed and the auto is deduced as int.
In a function declaration that uses the trailing return type syntax, the keyword auto does not perform automatic type detection. It only serves as a part of the syntax.
In a function declaration that does not use the trailing return type syntax, the keyword auto indicates that the return type will be deduced from the operand of its return statement using the rules for template argument deduction.
If the declared type of the variable is decltype(auto), the keyword auto is replaced with the expression (or expression list) of its initializer, and the actual type is deduced using the rules for decltype.
If the return type of the function is declared decltype(auto), the keyword auto is replaced with the operand of its return statement, and the actual return type is deduced using the rules for decltype.
A nested-name-specifier of the form auto:: is a placeholder that is replaced by a class or enumeration type following the rules for constrained type placeholder deduction.
A parameter declaration in a lambda expression. (since C++14) A function parameter declaration. (concepts TS)
Notes
Until C++11, auto had the semantic of a storage duration specifier.
Mixing auto variables and functions in one declaration, as in auto f() -> int, i = 0; is not allowed.
For more info : http://en.cppreference.com/w/cpp/language/auto
This functionality hasn't been there your whole life. It's been supported in Visual Studio since the 2010 version. It's a new C++11 feature, so it's not exclusive to Visual Studio and is/will be portable. Most compilers support it already.
The auto keyword is an important and frequently used keyword for C ++.When initializing a variable, auto keyword is used for type inference(also called type deduction).
There are 3 different rules regarding the auto keyword.
First Rule
auto x = expr; ----> No pointer or reference, only variable name. In this case, const and reference are ignored.
int y = 10;
int& r = y;
auto x = r; // The type of variable x is int. (Reference Ignored)
const int y = 10;
auto x = y; // The type of variable x is int. (Const Ignored)
int y = 10;
const int& r = y;
auto x = r; // The type of variable x is int. (Both const and reference Ignored)
const int a[10] = {};
auto x = a; // x is const int *. (Array to pointer conversion)
Note : When the name defined by auto is given a value with the name of a function,
the type inference will be done as a function pointer.
Second Rule
auto& y = expr; or auto* y = expr; ----> Reference or pointer after auto keyword.
Warning : const is not ignored in this rule !!! .
int y = 10;
auto& x = y; // The type of variable x is int&.
Warning : In this rule, array to pointer conversion (array decay) does not occur !!!.
auto& x = "hello"; // The type of variable x is const char [6].
static int x = 10;
auto y = x; // The variable y is not static.Because the static keyword is not a type. specifier
// The type of variable x is int.
Third Rule
auto&& z = expr; ----> This is not a Rvalue reference.
Warning : If the type inference is in question and the && token is used, the names
introduced like this are called "Forwarding Reference" (also called Universal Reference).
auto&& r1 = x; // The type of variable r1 is int&.Because x is Lvalue expression.
auto&& r2 = x+y; // The type of variable r2 is int&&.Because x+y is PRvalue expression.
The auto keyword specifies that the type of the variable that is being declared will be automatically deducted from its initializer. In case of functions, if their return type is auto then that will be evaluated by return type expression at runtime.
It can be very useful when we have to use the iterator. For e.g. for below code we can simply use the "auto" instead of writing the whole iterator syntax .
int main()
{
// Initialize set
set<int> s;
s.insert(1);
s.insert(4);
s.insert(2);
s.insert(5);
s.insert(3);
// iterator pointing to
// position where 2 is
auto pos = s.find(3);
// prints the set elements
cout << "The set elements after 3 are: ";
for (auto it = pos; it != s.end(); it++)
cout << *it << " ";
return 0;
}
This is how we can use "auto" keyword
It's not going anywhere ... it's a new standard C++ feature in the implementation of C++11. That being said, while it's a wonderful tool for simplifying object declarations as well as cleaning up the syntax for certain call-paradigms (i.e., range-based for-loops), don't over-use/abuse it :-)
It's Magic is it's ability to reduce having to write code for every Variable Type passed into specific functions. Consider a Python similar print() function in it's C base.
#include <iostream>
#include <string>
#include <array>
using namespace std;
void print(auto arg) {
cout<<arg<<" ";
}
int main()
{
string f = "String";//tok assigned
int x = 998;
double a = 4.785;
string b = "C++ Auto !";
//In an opt-code ASCII token stream would be iterated from tok's as:
print(a);
print(b);
print(x);
print(f);
}