Use auto for only one variable with structured binding

Question

I am trying to update a variable passed to a function with a structured binding:

#include <iostream>
#include <tuple>
#include <utility>

std::pair<int, double> func(double y)
{
    return {1, 1.3+y};
}

int main() {
    double y = 1.0;
    auto [x, y2] = func(y);
    y = y2;
    std::cout << "x = " << x << ", y = " << y << 'n';
    return 0;
}

Is it possible to avoid the extra assignment y = y2 and using something like

[auto x, y] = func(y); // this does not work

or

std::tie(auto x, y) = func(y); // this does not work

c++c++17

Marshall Clow · Accepted Answer

To answer your question narrowly:
No, you can't reuse an existing variable in a structured binding. Structured binding always declares new variables.
That being said, there are other options - like std::tie, as @Robert A has demonstrated.

dfrib · Answer

A structured binding is a declaration; it cannot be used e.g. for assignment into an already declared variable.
If you are allowed to move the declaration of y and you ever only need it to make a function call, you could abuse the scope of the the capture list of an immediately invoked lambda, and let it shadow (only within the scope of the lambda) the variable y that is declared as part of a structured binding, which is in turn initialized using the return from the immediately invoked lambda:
auto [x, y] = [y = 1.0](){ return func(y); }();
            // ^^^^^^^ actually not at all in namespace scope,
            //         but a data member of the closure type
            //         of the lambda expression.

You could likewise use a named lambda:
const auto l = [y = 1.0](){ return func(y); };
auto [x, y] = l();

As is typically the case with shadowing alongside the somewhat complex scoping rules of C++, this is likely only to confuse readers, though.

Robert Andrzejuk · Answer

A third option:
decltype(func(y).first) x;
std::tie(x, y) = func(y);

Use auto for only one variable with structured binding

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