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# return statement

From cppreference.com
< cpp‎ | language

Terminates current function and returns specified value to the caller function.

### [edit] Syntax

attr(optional) `return` expression(optional) `;` | (1) |
attr(optional) `return` braced-init-list `;` | (2) | (since C++11)

attr(C++11) | - | optional sequence of any number of attributes
expression | - | expression, convertible to the function return type
braced-init-list | - | brace-enclosed list of initializers and other braced-init-lists

### [edit] Explanation

1) Evaluates the expression, terminates the current function and returns the result of the expression to the caller, after implicit conversion to the function return type. The expression is optional in functions whose return type is (possibly cv-qualified) void, in constructors and in destructors.
2) Uses copy-list-initialization to construct the return value of the function.

### [edit] Notes

If control reaches the end of a function without encountering a return statement, return; is executed (except in the main function, where return 0; is executed).

Flowing off the end of a value-returning function (except main) without a return statement is undefined behavior.

In a function returning void, the return statement with expression can be used, if the expression type is void.

Returning by value may involve construction and copy/move of a temporary object, unless copy elision is used.

If expression is an lvalue expression and the conditions for copy elision are met, or would be met, except that expression names a function parameter, then overload resolution to select the constructor to use for initialization of the returned value is performed twice: first as if expression were an rvalue expression (thus it may select the move constructor or a copy constructor taking reference to const), and if no suitable conversion is available, overload resolution is performed the second time, with lvalue expression (so it may select the copy constructor taking a reference to non-const). | (since C++11)
The above rule applies even if the function return type is different from the type of expression (copy elision requires same type) | (since C++14)

### [edit] Keywords

`return`

### [edit] Example

Run this code

    #include <iostream>
    #include <string>
    #include <utility>

    void fa(int i)
    {
        if (i == 2)
             return;
        std::cout << i << '\n';
    } // implied return;

    int fb(int i)
    {
        if (i > 4)
             return 4;
        std::cout << i << '\n';
        return 2;
    }

    std::pair<std::string, int> fc(const char* p, int x)
    {
        return {p, x};
    }

    void fd()
    {
        return fa(10); // fa(10) is a void expression
    }

    int main()
    {
        fa(2); // returns, does nothing when i==2
        fa(1); // prints its argument, then returns
        int i = fb(5); // prints its argument, returns 2
        i = fb(i); // prints its argument, returns 4
        std::cout << i << '\n'
                  << fc("Hello", 7).second << '\n';;
        fd();
    }

Output:

    1
    4
    2
    7
    10

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