Difference between `constexpr` and `const`

2楼-- · 2018-12-31 10:41

According to book of "The C++ Programming Language 4th Editon" by Bjarne Stroustrup
• const: meaning roughly ‘‘I promise not to change this value’’ (§7.5). This is used primarily to specify interfaces, so that data can be passed to functions without fear of it being modified.
The compiler enforces the promise made by const.
• constexpr: meaning roughly ‘‘to be evaluated at compile time’’ (§10.4). This is used primarily to specify constants, to allow
For example:

const int dmv = 17; // dmv is a named constant
int var = 17; // var is not a constant
constexpr double max1 = 1.4*square(dmv); // OK if square(17) is a constant expression
constexpr double max2 = 1.4∗square(var); // error : var is not a constant expression
const double max3 = 1.4∗square(var); //OK, may be evaluated at run time
double sum(const vector<double>&); // sum will not modify its argument (§2.2.5)
vector<double> v {1.2, 3.4, 4.5}; // v is not a constant
const double s1 = sum(v); // OK: evaluated at run time
constexpr double s2 = sum(v); // error : sum(v) not constant expression

For a function to be usable in a constant expression, that is, in an expression that will be evaluated by the compiler, it must be defined constexpr.
For example:

constexpr double square(double x) { return x∗x; }

To be constexpr, a function must be rather simple: just a return-statement computing a value. A constexpr function can be used for non-constant arguments, but when that is done the result is not a constant expression. We allow a constexpr function to be called with non-constant-expression arguments in contexts that do not require constant expressions, so that we don’t hav e to define essentially the same function twice: once for constant expressions and once for variables.
In a few places, constant expressions are required by language rules (e.g., array bounds (§2.2.5, §7.3), case labels (§2.2.4, §9.4.2), some template arguments (§25.2), and constants declared using constexpr). In other cases, compile-time evaluation is important for performance. Independently of performance issues, the notion of immutability (of an object with an unchangeable state) is an important design concern (§10.4).

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柔情千种

3楼-- · 2018-12-31 10:42

A const int var can be dynamically set to a value at runtime and once it is set to that value, it can no longer be changed.

A constexpr int var cannot be dynamically set at runtime, but rather, at compile time. And once it is set to that value, it can no longer be changed.

Here is a solid example:

int main(int argc, char*argv[]) {
    const int p = argc; 
    // p = 69; // cannot change p because it is a const
    // constexpr int q = argc; // cannot be, bcoz argc cannot be computed at compile time 
    constexpr int r = 2^3; // this works!
    // r = 42; // same as const too, it cannot be changed
}

The snippet above compiles fine and I have commented out those that cause it to error.

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浪荡孟婆

4楼-- · 2018-12-31 10:43

Overview

const guarantees that a program does not change an object’s value. However, const does not guarantee which type of initialization the object undergoes.

Consider:
```
const int mx = numeric_limits<int>::max();  // OK: runtime initialization
```
The function max() merely returns a literal value. However, because the initializer is a function call, mx undergoes runtime initialization. Therefore, you cannot use it as a constant expression:
```
int arr[mx];  // error: “constant expression required”
```
constexpr is a new C++11 keyword that rids you of the need to create macros and hardcoded literals. It also guarantees, under certain conditions, that objects undergo static initialization. It controls the evaluation time of an expression. By enforcing compile-time evaluation of its expression, constexpr lets you define true constant expressions that are crucial for time-critical applications, system programming, templates, and generally speaking, in any code that relies on compile-time constants.

Constant-expression functions

A constant-expression function is a function declared constexpr. Its body must be non-virtual and consist of a single return statement only, apart from typedefs and static asserts. Its arguments and return value must have literal types. It can be used with non-constant-expression arguments, but when that is done the result is not a constant expression.

A constant-expression function is meant to replace macros and hardcoded literals without sacrificing performance or type safety.

constexpr int max() { return INT_MAX; }           // OK
constexpr long long_max() { return 2147483647; }  // OK
constexpr bool get_val()
{
    bool res = false;
    return res;
}  // error: body is not just a return statement

constexpr int square(int x)
{ return x * x; }  // OK: compile-time evaluation only if x is a constant expression
const int res = square(5);  // OK: compile-time evaluation of square(5)
int y = getval();
int n = square(y);          // OK: runtime evaluation of square(y)

Constant-expression objects

A constant-expression object is an object declared constexpr. It must be initialized with a constant expression or an rvalue constructed by a constant-expression constructor with constant-expression arguments.

A constant-expression object behaves as if it was declared const, except that it requires initialization before use and its initializer must be a constant expression. Consequently, a constant-expression object can always be used as part of another constant expression.

struct S
{
    constexpr int two();      // constant-expression function
private:
    static constexpr int sz;  // constant-expression object
};
constexpr int S::sz = 256;
enum DataPacket
{
    Small = S::two(),  // error: S::two() called before it was defined
    Big = 1024
};
constexpr int S::two() { return sz*2; }
constexpr S s;
int arr[s.two()];  // OK: s.two() called after its definition

Constant-expression constructors

A constant-expression constructor is a constructor declared constexpr. It can have a member initialization list but its body must be empty, apart from typedefs and static asserts. Its arguments must have literal types.

A constant-expression constructor allows the compiler to initialize the object at compile-time, provided that the constructor’s arguments are all constant expressions.

struct complex
{
    // constant-expression constructor
    constexpr complex(double r, double i) : re(r), im(i) { }  // OK: empty body
    // constant-expression functions
    constexpr double real() { return re; }
    constexpr double imag() { return im; }
private:
    double re;
    double im;
};
constexpr complex COMP(0.0, 1.0);         // creates a literal complex
double x = 1.0;
constexpr complex cx1(x, 0);              // error: x is not a constant expression
const complex cx2(x, 1);                  // OK: runtime initialization
constexpr double xx = COMP.real();        // OK: compile-time initialization
constexpr double imaglval = COMP.imag();  // OK: compile-time initialization
complex cx3(2, 4.6);                      // OK: runtime initialization

Tips from the book Effective Modern C++ by Scott Meyers about constexpr:

constexpr objects are const and are initialized with values known during compilation;
constexpr functions produce compile-time results when called with arguments whose values are known during compilation;
constexpr objects and functions may be used in a wider range of contexts than non-constexpr objects and functions;
constexpr is part of an object’s or function’s interface.

Source: Using constexpr to Improve Security, Performance and Encapsulation in C++.

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步步皆殇っ

5楼-- · 2018-12-31 10:47

const applies for variables, and prevents them from being modified in your code.

constexpr tells the compiler that this expression results in a compile time constant value, so it can be used in places like array lengths, assigning to const variables, etc. The link given by Oli has a lot of excellent examples.

Basically they are 2 different concepts altogether, and can (and should) be used together.

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何处买醉

6楼-- · 2018-12-31 10:50

Both const and constexpr can be applied to variables and functions. Even though they are similar to each other, in fact they are very different concepts.

Both const and constexpr mean that their values can't be changed after their initialization. So for example:

const int x1=10;
constexpr int x2=10;

x1=20; // ERROR. Variable 'x1' can't be changed.
x2=20; // ERROR. Variable 'x2' can't be changed.

The principal difference between const and constexpr is the time when their initialization values are known (evaluated). While the values of const variables can be evaluated at both compile time and runtime, constexpr are always evaluated at compile time. For example:

int temp=rand(); // temp is generated by the the random generator at runtime.

const int x1=10; // OK - known at compile time.
const int x2=temp; // OK - known only at runtime.
constexpr int x3=10; // OK - known at compile time.
constexpr int x4=temp; // ERROR. Compiler can't figure out the value of 'temp' variable at compile time so `constexpr` can't be applied here.

The key advantage to know if the value is known at compile time or runtime is the fact that compile time constants can be used whenever compile time constants are needed. For instance, C++ doesn't allow you to specify C-arrays with the variable lengths.

int temp=rand(); // temp is generated by the the random generator at runtime.

int array1[10]; // OK.
int array2[temp]; // ERROR.

So it means that:

const int size1=10; // OK - value known at compile time.
const int size2=temp; // OK - value known only at runtime.
constexpr int size3=10; // OK - value known at compile time.


int array3[size1]; // OK - size is known at compile time.
int array4[size2]; // ERROR - size is known only at runtime time.
int array5[size3]; // OK - size is known at compile time.

So const variables can define both compile time constants like size1 that can be used to specify array sizes and runtime constants like size2 that are known only at runtime and can't be used to define array sizes. On the other hand constexpr always define compile time constants that can specify array sizes.

Both const and constexpr can be applied to functions too. A const function must be a member function (method, operator) where application of const keyword means that the method can't change the values of their member (non-static) fields. For example.

class test
{
   int x;

   void function1()
   {
      x=100; // OK.
   }

   void function2() const
   {
      x=100; // ERROR. The const methods can't change the values of object fields.
   }
};

A constexpr is a different concept. It marks a function (member or non-member) as the function that can be evaluated at compile time if compile time constants are passed as their arguments. For example you can write this.

constexpr int func_constexpr(int X, int Y)
{
    return(X*Y);
}

int func(int X, int Y)
{
    return(X*Y);
}

int array1[func_constexpr(10,20)]; // OK - func_constexpr() can be evaluated at compile time.
int array2[func(10,20)]; // ERROR - func() is not a constexpr function.

int array3[func_constexpr(10,rand())]; // ERROR - even though func_constexpr() is the 'constexpr' function, the expression 'constexpr(10,rand())' can't be evaluated at compile time.

By the way the constexpr functions are the regular C++ functions that can be called even if non-constant arguments are passed. But in that case you are getting the non-constexpr values.

int value1=func_constexpr(10,rand()); // OK. value1 is non-constexpr value that is evaluated in runtime.
constexpr int value2=func_constexpr(10,rand()); // ERROR. value2 is constexpr and the expression func_constexpr(10,rand()) can't be evaluated at compile time.

The constexpr can be also applied to the member functions (methods), operators and even constructors. For instance.

class test2
{
    static constexpr int function(int value)
    {
        return(value+1);
    }

    void f()
    {
        int x[function(10)];


    }
};

A more 'crazy' sample.

class test3
{
    public:

    int value;

    // constexpr const method - can't chanage the values of object fields and can be evaluated at compile time.
    constexpr int getvalue() const
    {
        return(value);
    }

    constexpr test3(int Value)
        : value(Value)
    {
    }
};


constexpr test3 x(100); // OK. Constructor is constexpr.

int array[x.getvalue()]; // OK. x.getvalue() is constexpr and can be evaluated at compile time.

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栀子花@的思念

7楼-- · 2018-12-31 10:52

Basic meaning and syntax

Both keywords can be used in the declaration of objects as well as functions. The basic difference when applied to objects is this:

const declares an object as constant. This implies a guarantee that, once initialized, the value of that object won't change, and the compiler can make use of this fact for optimizations. It also helps prevent the programmer from writing code that modifies objects that were not meant to be modified after initialization.
constexpr declares an object as fit for use in what the Standard calls constant expressions. But note that constexpr is not the only way to do this.

When applied to functions the basic difference is this:

const can only be used for non-static member functions, not functions in general. It gives a guarantee that the member function does not modify any of the non-static data members.
constexpr can be used with both member and non-member functions, as well as constructors. It declares the function fit for use in constant expressions. The compiler will only accept it if the function meets certain criteria (7.1.5/3,4), most importantly ^(†):
- The function body must be non-virtual and extremely simple: Apart from typedefs and static asserts, only a single return statement is allowed. In the case of a constructor, only an initialization list, typedefs and static assert are allowed. (= default and = delete are allowed, too, though.)
- As of C++14 the rules are more relaxed, what is allowed since then inside a constexpr function: asm declaration, a goto statement, a statement with a label other than case and default, try-block, definition of a variable of non-literal type, definition of a variable of static or thread storage duration, definition of a variable for which no initialization is performed.
- The arguments and the return type must be literal types (i.e., generally speaking, very simple types, typically scalars or aggregates)

Constant expressions

As said above, constexpr declares both objects as well as functions as fit for use in constant expressions. A constant expression is more than merely constant:

It can be used in places that require compile-time evaluation, for example, template parameters and array-size specifiers:

template<int N>
class fixed_size_list
{ /*...*/ };

fixed_size_list<X> mylist;  // X must be an integer constant expression

int numbers[X];  // X must be an integer constant expression

But note:
- Declaring something as constexpr does not necessarily guarantee that it will be evaluated at compile time. It can be used for such, but it can be used in other places that are evaluated at run-time, as well.
- An object may be fit for use in constant expressions without being declared constexpr. Example:
```
int main()
{
  const int N = 3;
  int numbers[N] = {1, 2, 3};  // N is constant expression
}
```
This is possible because N, being constant and initialized at declaration time with a literal, satisfies the criteria for a constant expression, even if it isn't declared constexpr.

So when do I actually have to use constexpr?

An object like N above can be used as constant expression without being declared constexpr. This is true for all objects that are:
- const
- of integral or enumeration type and
- initialized at declaration time with an expression that is itself a constant expression
_{[This is due to §5.19/2: A constant expression must not include a subexpressions that involves "an lvalue-to-rvalue modification unless […] a glvalue of integral or enumeration type […]" Thanks to Richard Smith for correcting my earlier claim that this was true for all literal types.]}

For a function to be fit for use in constant expressions, it must be explicitly declared constexpr; it is not sufficient for it merely to satisfy the criteria for constant-expression functions. Example:

template<int N>
class list
{ };

constexpr int sqr1(int arg)
{ return arg * arg; }

int sqr2(int arg)
{ return arg * arg; }

int main()
{
  const int X = 2;
  list<sqr1(X)> mylist1;  // OK: sqr1 is constexpr
  list<sqr2(X)> mylist2;  // wrong: sqr2 is not constexpr
}

When can I / should I use both, const and constexpr together?

A. In object declarations. This is never necessary when both keywords refer to the same object to be declared. constexpr implies const.

constexpr const int N = 5;

is the same as

constexpr int N = 5;

However, note that there may be situations when the keywords each refer to different parts of the declaration:

static constexpr int N = 3;

int main()
{
  constexpr const int *NP = &N;
}

Here, NP is declared as an address constant-expression, i.e. an pointer that is itself a constant expression. (This is possible when the address is generated by applying the address operator to a static/global constant expression.) Here, both constexpr and const are required: constexpr always refers to the expression being declared (here NP), while const refers to int (it declares a pointer-to-const). Removing the const would render the expression illegal (because (a) a pointer to a non-const object cannot be a constant expression, and (b) &N is in-fact a pointer-to-constant).

B. In member function declarations. In C++11, constexpr implies const, while in C++14 and C++17 that is not the case. A member function declared under C++11 as

constexpr void f();

needs to be declared as

constexpr void f() const;

under C++14 in order to still be usable as a const function.

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