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Jacob Jingle

Returning tuple-local variable

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Jacob Jingle    226

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?
}

 

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Cornstalks    7030

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.

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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?

Edited by Servant of the Lord

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Cornstalks    7030

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:
std::tuple<std::array<int, 2>&, std::array<int, 2>&> // references!
 
However, that's not what the function returns. The function returns:
std::tuple<std::array<int, 2>, std::array<int, 2>>
 
So a copy must be made, much like if you did:
int foo()
{
    int x = 2;
    int& y = x;
 
    return y; // still safe, despite y being a reference, because the value gets copied
}
 
Indeed, we can try:
 
#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;
}
 
Running this, I got:

0x7fff54c9a7f8    0x7fff54c9a7f8
0x7fff54c9a7f0    0x7fff54c9a7f0
0x7fff54c9ab08
0x7fff54c9ab08    0x7fff54c9ab10
0x7fff54c9ab08    0x7fff54c9ab10
 
As you can see, the addresses in foo() aren't the same addresses as for t. At least as far as I've understood this so far.

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