10.9 — Type deduction for functions

Consider the following program:

int add(int x, int y)
{
    return x + y;
}

When this function is compiled, the compiler will determine that x + yevaluates to an int, then ensure that type of the return value matches the declared return type of the function (or that the return value type can be converted to the declared return type).

Since the compiler already has to deduce the return type from the return statement (to ensure that the value can be converted to the function’s declared return type), in C++14, the auto keyword was extended to do function return type deduction. This works by using the auto keyword in place of the function’s return type.

For example:

auto add(int x, int y)
{
    return x + y;
}

Because the return statement is returning an int value, the compiler will deduce that the return type of this function is int.

When using an auto return type, all return statements within the function must return values of the same type, otherwise an error will result. For example:

auto someFcn(bool b)
{
    if (b)
        return 5; // return type int
    else
        return 6.7; // return type double
}

In the above function, the two return statements return values of different types, so the compiler will give an error.

If such a case is desired for some reason, you can either explicitly specify a return type for your function (in which case the compiler will try to implicitly convert any non-matching return expressions to the explicit return type), or you can explicitly convert all of your return statements to the same type. In the example above, the latter could be done by changing 5 to 5.0, but static_cast can also be used for non-literal types.

Benefits of type deduction for functions

The biggest advantage of type deduction for functions is that having the compiler deduce the function’s return type negates the risk of a mismatched return type (preventing unexpected conversions).

In certain cases, the return type of a function may either be long and complex, or not be that obvious. In such cases, auto can be used to simplify:

// let compiler determine the return type of unsigned short + char
auto add(unsigned short x, char y)
{
    return x + y;
}

Once we cover function templates, we’ll see more examples where using an auto return type is useful (see lesson 11.8 -- Function templates with multiple template types for an example).

Downsides of type deduction for functions

There are two major downsides to type deduction to functions:

1. Functions that use an auto return type must be fully defined before they can be used (a forward declaration is not sufficient). For example:

#include <iostream>

auto foo();

int main()
{
    std::cout << foo() << '\n'; // the compiler has only seen a forward declaration at this point

    return 0;
}

auto foo()
{
    return 5;
}

On the author’s machine, this gives the following compile error:

error C3779: 'foo': a function that returns 'auto' cannot be used before it is defined.

This makes sense: a forward declaration does not have enough information for the compiler to deduce the function’s return type. This means normal functions that return auto are typically only callable from within the file in which they are defined.

2. When using type deduction with objects, the initializer is always present as part of the same statement, so it’s usually not overly burdensome to determine what type will be deduced. With type deduction for functions, that is not the case -- the function’s prototype gives no indication as to what type the function actually returns. A good programming IDE should make clear what the deduced type of the function is, but in absence of having that available, a user would actually have to dig into the function body itself to determine what type the function returned. The odds of mistakes being made are higher.

Unlike type deduction for objects, there isn’t as much consensus on best practices for function return type deduction. We recommend avoiding type deduction for normal functions.

Trailing return type syntax

The auto keyword can also be used to declare functions using a trailing return syntax, where the return type is specified after the rest of the function prototype.

Consider the following function:

int add(int x, int y)
{
  return (x + y);
}

Using the trailing return syntax, this could be equivalently written as:

auto add(int x, int y) -> int
{
  return (x + y);
}

In this case, auto does not perform type deduction -- it is just part of the syntax to use a trailing return type.

Why would you want to use this? Here are some reasons:

1. For functions with complex return types, a trailing return type can make the function easier to read:

#include <type_traits> // for std::common_type

std::common_type_t<int, double> compare(int, double);         // harder to read (where is the name of the function in this mess?)
auto compare(int, double) -> std::common_type_t<int, double>; // easier to read (we don't have to read the return type unless we care)

2. The trailing return type syntax can be used to align the names of your functions, which makes consecutive function declarations easier to read:

auto add(int x, int y) -> int;
auto divide(double x, double y) -> double;
auto printSomething() -> void;
auto generateSubstring(const std::string &s, int start, int len) -> std::string;

#include <type_traits>
// note: decltype(x) evaluates to the type of x

std::common_type_t<decltype(x), decltype(y)> add(int x, double y);         // Compile error: compiler hasn't seen definitions of x and y yet
auto add(int x, double y) -> std::common_type_t<decltype(x), decltype(y)>; // ok

For now, we recommend the continued use of the traditional function return syntax except in situations that require the trailing return syntax.

Type deduction can’t be used for function parameter types

Many new programmers who learn about type deduction try something like this:

#include <iostream>

void addAndPrint(auto x, auto y)
{
    std::cout << x + y << '\n';
}

int main()
{
    addAndPrint(2, 3); // case 1: call addAndPrint with int parameters
    addAndPrint(4.5, 6.7); // case 2: call addAndPrint with double parameters

    return 0;
}

Unfortunately, type deduction doesn’t work for function parameters, and prior to C++20, the above program won’t compile (you’ll get an error about function parameters not being able to have an auto type).

In C++20, the auto keyword was extended so that the above program will compile and function correctly -- however, auto is not invoking type deduction in this case. Rather, it is triggering a different feature called function templates that was designed to actually handle such cases.