3

I have to work with some heavily templated code which I am trying to decipher.

I see two different constructs and I am not sure if I am missing something. Here is a simplified example of these type conversions which are used at some places, is there a difference between the following statements?

template<typename T, typename S> S my_function(T t) 
{
    // version 1:
    S s = t
    return s;

    // version 2:
    return S(t);
 }

I am thinking to changing everything to one style, are the two statements 100% equivalent and if not, what are the differences?

Beginner
  • 5,277
  • 6
  • 34
  • 71
  • 1
    Use `function` as the name is a bad practice. – iBug Jan 31 '18 at 11:23
  • With compiler optimization, I think so, but am not 100% sure. – iBug Jan 31 '18 at 11:24
  • 1
    You probably want to swap the template parameter positions for `T` and `S` so you can use `function` with `T` deduced instead of having to write `function`. – nwp Jan 31 '18 at 11:27
  • 2
    For `T == int *` and `S == double *` version 1 will fail because incompatible types and version 2 will compile because C cast. – nwp Jan 31 '18 at 11:28
  • 1
    I feel the answer can also vary greatly between [tag:c++17] and [tag:c++14]. A proper tagging may help focus the effort in answering. – StoryTeller - Unslander Monica Jan 31 '18 at 11:39
  • @StoryTeller A complete answer would be a bunch of if-elses. So I just throw in the relevant cppref pages for the OP to figure out the if-elses. – Lingxi Jan 31 '18 at 11:50

2 Answers2

3

No, they are not 100% equivalent!

For the simple case (c++11 and beyond), considering the return statement with type class S then the difference are:

If S has move operators then they are used when returning the value, otherwise the copy operators are used. Also note that if those operators are private or deleted then version 1 will not compile.

For version 2 s is created (with a constructor, or user defined operator) and returned directly because of RVO

Now for the complicated case, there are other things to consider; First version uses copy initialization and second is direct initialization. There are also user defined operators and explicit keyword that can alter the behavior.

Raxvan
  • 6,257
  • 2
  • 25
  • 46
1

Version 1 and 2 are called copy initialization (form 1) and direct initialization (form 3), respectively. The difference depends on the relationship between T and S.

The effects on the return statement, which performs copy initialization (form 4), depend on the nature of S. The difference is that s is an l-value while S(t) is an r-value. Copy elision may also apply here. For class types, this may make a difference.

Lingxi
  • 14,579
  • 2
  • 37
  • 93