[PDF] constexpr lambda capture

  • Are lambda captures const in C++?

    Typically, a lambda's function call operator is const-by-value, but use of the mutable keyword cancels this out.
    It doesn't produce mutable data members.
    The mutable specification enables the body of a lambda expression to modify variables that are captured by value.

  • Why is lambda capture const?

    By default, variables are captured by const value .
    This means when the lambda is created, the lambda captures a constant copy of the outer scope variable, which means that the lambda is not allowed to modify them.
    In the following example, we capture the variable ammo and try to decrement it.

  • What is lambda capture C++?

    A capture clause of lambda definition is used to specify which variables are captured and whether they are captured by reference or by value.
    An empty capture closure [ ], indicates that no variables are used by lambda which means it can only access variables that are local to it.

  • What is lambda capture C++?

    Capture clauses are used to instruct the compiler to capture the variables visible in scope around where the lambda is declared, either by copy or by reference, to be available for use inside the lambda method body.
    Capture clauses can be used in different ways to specify the requirements of functions.

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Wording for Constexpr Lambda

2016?3?1? constexpr function (similar to the constexpr inference that ... For a generic lambda with no lambda-capture the closure type has a public.



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2015?4?28? It seems that constexpr lambdas aren't inherently wrong so ... all of its data members (that correspond to each capture) are core constant.



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2017?2?5? During discussion of constexpr if (P0252) at Oulu a particularly problematic case was discovered: template<typename T> void f(T t) {.



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2018?5?29? and C++17 to capture modes in part 1 of this series [1] this report will look ... Listing 11: Explicitly constexpr lambda expression.



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Everything you (n)ever wanted to know about C++s Lambdas

…no capturing takes no parameters and returns nothing [=]: captures all variables (used in the lambda) by value ... lambdas are default constexpr.



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C++ Technical Specification — Extensions for Reflection

2018?8?11? 1 In C++ [expr.prim.lambda.capture] apply the following change to paragraph 7: ... constexpr auto is_public_v = is_public<T>::value;.

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