Is there a way to pass auto as an argument in C++?

Question

Is there a way to pass auto as an argument to another function?

int function(auto data) {     //DOES something } 

Answer 1

C++20 allows auto as function parameter type

This code is valid using C++20:

int function(auto data) {    // do something, there is no constraint on data } 

As an abbreviated function template.

This is a special case of a non constraining type-constraint (i.e. unconstrained auto parameter). Using concepts, the constraining type-constraint version (i.e. constrained auto parameter) would be for example:

void function(const Sortable auto& data) {     // do something that requires data to be Sortable     // assuming there is a concept named Sortable } 

The wording in the spec, with the help of my friend Yehezkel Bernat:

9.2.8.5 Placeholder type specifiers [dcl.spec.auto]

placeholder-type-specifier:

type-constraint_opt auto

type-constraint_opt decltype ( auto )

1. A placeholder-type-specifier designates a placeholder type that will be replaced later by deduction from an initializer.

2. A placeholder-type-specifier of the form type-constraint_opt auto can be used in the decl-specifier-seq of a parameter-declaration of a function declaration or lambda-expression and signifies that the function is an abbreviated function template (9.3.3.5) ...

Answer 2

If you want that to mean that you can pass any type to the function, make it a template:

template <typename T> int function(T data); 

There's a proposal for C++17 to allow the syntax you used (as C++14 already does for generic lambdas), but it's not standard yet.

Edit: C++ 2020 now supports auto function parameters. See Amir's answer below