I'm kindof confused over the std::tie function. Is my return code safe? Or is this a local variable error?
std::tuple<std::array<int, 2>, std::array<int, 2>> FunctionThatReturnsTuple()
{
std::array<int, 2> buff = {1, 2};
std::array<int, 2> buff2 = {1, 2};
return std::tie(buff, buff2); <---- safe?
}
It's okay because it's safe to return std::arrays from a function (unlike a "raw" array like int b[2];). std::array is treated like a "class" and its data is copied. Imagine the following code:
struct Foo
{
int a[10];
};
Foo function()
{
Foo f;
return f; // Totally safe because of Foo's struct semantics
}
Whether it's in a tuple or not doesn't matter here.
I'm new to tuples, so this question is as much for my own education than as an answer to your question, but shouldn't that be std::make_tuple() not std::tie()?
Reading the documentation I just linked to, it says std::tie() makes a std::tuple of references. Once your std::array goes out of scope, wouldn't those references would be invalid?
Whereas std::make_tuple() copies the values, so the tuple owns it's own arrays.
To get rid of the copy, you could even use std::move():
return std::make_tuple(std::move(buff), std::move(buff2)); //I don't know if the std::move()s are done automatically be the compiler.
This is assuming that 'buff' and 'buff2' are never used again, like in your example.
If the two std::arrays continue to exist after the function ends, say, as member-variables of a class, then the std::tie() version that takes references would be valid as long as the class holding the real value is valid.
Am I missing something here?
It's entirely possible I missed something in my first post and I'll have to embarrassingly revoke it, but my understanding is that in this example, std::tie results in:Am I missing something here?
std::tuple<std::array<int, 2>&, std::array<int, 2>&> // references!
std::tuple<std::array<int, 2>, std::array<int, 2>>
int foo()
{
int x = 2;
int& y = x;
return y; // still safe, despite y being a reference, because the value gets copied
}
#include <typeinfo>
#include <iostream>
#include <tuple>
#include <array>
std::tuple<std::array<int, 2>, std::array<int, 2>> foo()
{
std::array<int, 2> buff = {1, 2};
std::array<int, 2> buff2 = {1, 2};
std::cout << &buff << '\t' << buff.data() << std::endl;
std::cout << &buff2 << '\t' << buff2.data() << std::endl;
return std::tie(buff, buff2);
}
int main()
{
auto t = foo();
std::cout << &t << std::endl;
std::cout << &std::get<0>(t) << '\t' << &std::get<1>(t) << std::endl;
std::cout << std::get<0>(t).data() << '\t' << std::get<1>(t).data() << std::endl;
}
Ah, forgot about the return value being converted - thanks for the clarification!
Had I remembered it, I would've guessed that it wouldn't have compiled anyway. 
