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3 Answers Sorted by: 32 Yes; std::tie was invented for this: #include
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Arrays: Let's try to bind to an array: int an_array[] = {1337, 1337}; auto& [x, y] = an_array; x = 42; y = 42; std::cout << an_array[0] << "," << an_array[1]; // What happen here? After substitution of [x, y] with our anonymous variable we obtain:
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const auto [x, y] = FuncReturningStdPair (); Description: Allows writing auto [x, y, z] = expr; where the type of expr is a tuple-like object, whose elements are bound to the variables x , y , and z (which this construct declares).
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Auto Type Deduction. Just to refresh your memory, when you use auto, the compiler determines the type that will replace the auto keyword by using the rules for template type deduction from a function call. Among other things, this basically means that when you write a bare auto, you get a copy: int x = 0; auto y = x; y = 1;
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Type deduction (also sometimes called type inference) is a feature that allows the compiler to deduce the type of an object from the object's initializer.To use type deduction, the auto keyword is used in place of the variable's type:. int main() { auto d{ 5.0 }; // 5.0 is a double literal, so d will be type double auto i{ 1 + 2 }; // 1 + 2 evaluates to an int, so i will be type int auto x.
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In addition to non-dependent packs, this paper also seems like it would offer the ability to declare a pack at namespace scope: struct Point { int x, y; }; auto [. parts] = Point{1, 2}; Structured bindings in namespace scope are a little odd to begin with, since they currently cannot be declared inline.
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x binds to int stored in the generated object that is a copy of a. And similarly, y binds to float. Modifiers Several modifiers can be used with structured bindings (similarly like on auto ): const modifiers: const auto [a, b, c,.] = expression; References: auto& [a, b, c,.] = expression; auto&& [a, b, c,.] = expression; For example:
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auto : auto ref operator : either & or && identifier-list : List of comma separated variable names. expression : An expression that does not have the comma operator at the top level (i.e, an assignment-expression), and has either array or non-union class type. Basic Type Deduction : Let E denote the type of the initializer expression.
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