In this code:
int a=8;
const int &b = a; //b is a Low-Level const
auto c = b;
I expected c to be constant int type since it has been autoed to a low level const. But it is referring to a (which has been aliased by b) instead if b for its declaration.
c picks the b's type only when it is used like auto &c =b
What I understand is only top-level const is dropped. So can someone please explain what is wrong with my understanding?
Whereas in case of a pointer, the low level const is picked by auto:
int i=9;
const int * p=&i;
auto *d=p;
*d=45; // error: assignment of read-only location ‘* d’
.
The type of an expression is never a reference type. Expressions have non-reference types, and value categories.
The declaration const int &b = a; means:
There will be a variable called b.
The expression b has type const int.
The expression b designates the same object as a. In other words, the names a and b both refer to the same object.
The code decltype(b) resolves to const int&, but apart from that, you can't access the "history" of a reference. Once it's been bound, it's just a name for an int thereafter.
auto c = X; means that c is not declared as a reference; and its type is deduced from the the type of the expression X. If X happens to be the identifier of a variable, that does NOT invoke any special rules; X is treated as an expression in all cases.
The syntax decltype(auto) C = X; exists; if X is the name of a variable then this means that C is declared in the same way as X. (If X is not the name of a variable then the behaviour depends on the value category of X too).
In your first example, c is deduced from the type of the expression b which is const int. As you noted, top-level const is dropped here, so auto c means int c.
In your second example, the type of the expression p is const int *. This is not a top-level const, so auto d = p; would be const int * d = p;. In this case auto *d is not different to auto d.
Auto uses the template type deduction rules. The long and the short of it is that references get dropped when using auto (which is the reason you'll see things like decltype(auto) as returns for functions, as this uses a different set of rules for type deduction).
Scott Meyers goes through the rules in some detail in Item 2 of Effective Modern C++.
In the first example b and a are linked however c becomes the constant, this is because of the const in b taking effect. The problem with the second example is that *d is type const int which cannot be modified.
Hope this helps.
Related
I know I can do
auto&& bla = something();
and depending on the constness of the return value of something, I'd get a different type for bla.
Does this also work in the structured bindings case, e.g.
auto&& [bla, blabla] = something();
I would guess so (structured bindings piggy-back on auto initializers, which behave like this), but I can't find a definitive yes.
Update: Preliminary tests seem to do what I expect (derive the constness properly):
#include <tuple>
using thing = std::tuple<char, int*, short&, const double, const float&>;
int main()
{
char c = 0;
int i = 1;
short s = 2;
double d = 3.;
float f = 4.f;
thing t{c, &i, s, d, f};
auto&& [cc, ii, ss, dd, ff] = t;
c = 10;
*ii = 11;
ss = 12;
dd = 13.;
ff = 14.f;
}
Live demo, gives error as I'd expect if auto&& is doing its job:
main.cpp: In function 'int main()':
main.cpp:20:10: error: assignment of read-only reference 'dd'
dd = 13.;
^~~
main.cpp:21:10: error: assignment of read-only reference 'ff'
ff = 14.f;
I'd still like to know exactly where this behaviour is specified.
Note: Using "forwarding references" to mean this behaviour might be stretching it, but I don't have a good name to give the const deduction part of auto&& (or template-T&& for that matter).
Yes. Structured bindings and forwarding references mix well†.
In general, any place‡ you can use auto, you can use auto&& to acquire the different meaning. For structured bindings specifically, this comes from [dcl.struct.bind]:
Otherwise, e is defined as-if by
attribute-specifier-seqopt decl-specifier-seq ref-qualifieropt e initializer ;
where the declaration is never interpreted as a function declaration and the parts of the declaration other than the declarator-id are taken from the corresponding structured binding declaration.
There are further restrictions on these sections in [dcl.dcl]:
A simple-declaration with an identifier-list is called a structured binding declaration ([dcl.struct.bind]). The decl-specifier-seq shall contain only the type-specifier auto and cv-qualifiers. The initializer shall be of the form “= assignment-expression”, of the form “{ assignment-expression }”, or of the form “( assignment-expression )”, where the assignment-expression is of array or non-union class type.
Putting it together, we can break down your example:
auto&& [bla, blabla] = something();
as declaring this unnamed variable:
auto && e = something();
~~~~ ~~ ~~~~~~~~~~~
decl-specifier-seq initializer
ref-qualifier
The behavior is that is derived from [dcl.spec.auto] (specifically here). There, we do do deduction against the initializer:
template <typename U> void f(U&& );
f(something());
where the auto was replaced by U, and the && carries over. Here's our forwarding reference. If deduction fails (which it could only if something() was void), our declaration is ill-formed. If it succeeds, we grab the deduced U and treat our declaration as if it were:
U&& e = something();
Which makes e an lvalue or rvalue reference, that is const qualified for not, based on the value category and type of something().
The rest of the structured bindings rules follow in [dcl.struct.bind], based on the underlying type of e, whether or not something() is an lvalue, and whether or not e is an lvalue reference.
† With one caveat. For a structured binding, decltype(e) always is the referenced type, not the type you might expect it be. For instance:
template <typename F, typename Tuple>
void apply1(F&& f, Tuple&& tuple) {
auto&& [a] = std::forward<Tuple>(tuple);
std::forward<F>(f)(std::forward<decltype(a)>(a));
}
void foo(int&&);
std::tuple<int> t(42);
apply1(foo, t); // this works!
I pass my tuple is an lvalue, which you'd expect to pass its underlying elements in as lvalue references, but they actually get forwarded. This is because decltype(a) is just int (the referenced type), and not int& (the meaningful way in which a behaves). Something to keep in mind.
‡ There are two places I can think of where this is not the case.
In trailing-return-type declarations, you must use just auto. You can't write, e.g.:
auto&& foo() -> decltype(...);
The only other place I can think of where this might not be the case is part of the Concepts TS where you can use auto in more places to deduce/constrain types. There, using a forwarding reference when the type you're deducing isn't a reference type would be ill-formed I think:
std::vector<int> foo();
std::vector<auto> a = foo(); // ok, a is a vector<int>
std::vector<auto&&> b = foo(); // error, int doesn't match auto&&
const int ci = 10;
auto i = ci; // i will be "int" instead of "const int"
i = 20;
I am wondering why auto is designed for this kind of behaviour?
why the type i is "int" instead of "const int" ?
what is the concern here?
I think understand why will help us to remember it
auto mostly follows the same type deduction rules as template argument deduction. The only difference is that auto will deduce std::initializer_list from a braced-init-list in some cases, while template argument deduction doesn't do this.
From N3337, §7.1.6.4 [dcl.spec.auto]
6 ... The type deduced for the variable d is then
the deduced A determined using the rules of template argument deduction from a function call (14.8.2.1), ...
The behavior you're observing is the same as what template argument deduction would do when deducing types from a function call
§14.8.2.1 [temp.deduct.call]
2 If P is not a reference type:
— ...
— If A is a cv-qualified type, the top level cv-qualifiers of A’s type are ignored for type deduction.
Thus, in
auto i = ci;
the top level const qualifier is ignored and i is deduced as int.
When you write
auto& i = ci;
then i is no longer not a reference type and the above rule doesn't apply, so the const qualifier is retained.
auto by itself means that you want a new, locally-owned variable with a copy of the given value. const-ness is not part of value. An int is an int whether it's specified using a literal, a named constant, an expression, or a non-const variable.
auto i = 3,
j = i,
k = ci,
m = 3 + 4; // All these variables are type int.
To get a constant of deduced type, you can still use auto const. This expresses within the declaration how the variable may be used.
const auto i = 3;
Since C++14, there is also the decltype(auto) specifier which applies decltype to the initializer, to make a carbon copy of the given variable. Perhaps that's really what you expected:
decltype(auto) i = ci; // i receives type const int.
Live demo.
decltype(auto) is a bit tricky, though, and it has few use cases aside from its original purpose relating to deciding the return type of function call wrappers. Unless there's a good reason, choose const auto or const int instead.
Another alternative is to use a forwarding reference, spelled auto &&. This refers to the variable or value that initializes it, whatever that may be.
auto && i = ci; // i receives type const int & and aliases ci.
This is less expressive and specific, but reliably declares i as an alias to ci. The other thing you tried was auto &, which is similar but only allows forming a reference to a preexisting variable.
auto & i = ci; // i receives type const int & and aliases ci.
A reference to a const int variable must be of type const int &, because otherwise it would permit illegal modification.
I see decltype(x) used inside macros where x is a variable name because the type of the object isn't known inside macros.
For example:
decltype(x) y = expr;
I could just have easily use auto instead of decltype. So what are those situations where decltype is needed for a variable type declaration instead of auto?
decltype becomes handy when you need to return some unknown type, which is evaluated during compilation:
template<class A, class B>
void MultiplyAB(A a, B b, decltype(a*b)& output)
{
output = a * b;
}
Additionally, if you don't like the way the output is handled by a reference, then you can also use the late-specified return type (and also use the decltype):
template<class A, class B>
auto MultiplyAB(A a, B b) -> decltype(a*b)
{
return a * b;
}
All of this, and more, is described by B. Stroustrup in the C++ FAQ.
You should use it when the required type of y is:
different (or potentially different) from the type of expr. If it was the same then auto would be more concise.
similarly for auto & or other modifications of the type of expr that auto can express.
and one of the following:
dependent on something in the surrounding code (i.e. not always the same type) and difficult to write using type traits or similar. This will tend to happen in template code. There might be a type trait that you can use to get the required type from the template parameters, but then again there might not so a use of decltype would save you defining one.
always the same type, (or dependent on template parameters in a way that is easy to express using existing type traits or similar) but the type is very long-winded to write and there is a much shorter and clear expression you can use instead.
So for example replacing std::iterator_traits<RandomAccessIterator>::value_type with decltype(*it) might well be a win, although auto does often handle such cases.
Subjective judgements enter at the point of "what is difficult", "what is long-winded" and "what is clear", but the rules of procedure can be the same regardless of how you make those judgements in specific cases.
When you want y to always have whatever the declared type of x is.
In the context of your question,
You should use decltype when you want a new variable with precisely the same type as the original variable.
You should use auto when you want to assign the value of some expression to a new variable and you want or need its type to be deduced.
decltype(x) y always declares y with precisely the same type as the type x was declared with. In particular:
If x has type const int then y will have type const int.
If x has type int[100] then y will have type int[100].
If x has type int f(int) then y will have type int f(int). Yes, this actually declares another function with the same type as the original.
If x has type int& then y will have type int&; and if x has type int&& then y will have type int&&.
auto y = x will declare y with the following types, when x has the following types:
If x has type const int, then y will have type int. That is, auto strips top-level cv-qualifiers.
If x has type int[100], then y will have type int*. That is, auto performs array to pointer conversion. [1]
If x has type int f(int), then y will have type int (*)(int). That is, auto performs function to function pointer conversion. [2]
Finally, if x has type int& or int&&, then y will have type int. That is, auto removes references.
[1] You can't use decltype here because you can't copy-initialize an array.
[2] You can't use decltype here because you can't initialize a function.
[3] The reason why auto strips references is that C++ has no expressions of reference type! Once initialized, the "reference ness" of a reference becomes invisible.
Note that decltype also does something entirely different when its argument is not an id-expression, which I won't get into here.
Whenever your variable type isn't related to the expression being evaluated.
E.g:
struct Bar
{
Bar(int) {} // implicitly constructable
}
struct Bar2
{
Bar2(int) {} // implicitly constructable
}
struct Foo
{
static Bar var;
}
struct Foo2
{
static Bar2 var;
}
template <typename T>
void dummy()
{
decltype(T::var) myVar = 42;
}
dummy<Foo>(); // myVar is of type Bar1
dummy<Foo2>(); // myVar is of type Bar2
auto myAutoVar = 42; // type is int
Of course this is just one use case, there are many more out there.
decltype is significantly more versatile that auto and can always be used in place of it. Therefore I think it's pretty safe to say that decltype should only be used in cases where it's completely necessary, so if auto produces the wrong result you should use decltype. Also you can't as of yet use auto in return types and parameters, so you can use decltype there as well. C++14 will significantly increase the potential uses of auto and I would guess c++17 will go further. So the situations to use decltype will only be when you need to change the resulting type of expr
Another thing to consider is that decltype isn't really necessary unless you're writing library code, auto is nice for everyday programming if you want to make your code more concise, it's up for debate wether using as much auto as possible is good, but it's virtually necessary when working with unutterable types like lambdas.
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);
}