Why can't I pass function evaluation as constructor parameter in c++?

Question

I was wondering why I can't pass a function evaluation as constructor parameter, i.e.:

A a(fstream(argv[1]))

if I try to call a.dump(), I get an error.

If I use

fsrteam fin(argv[1]);
A a(fin);

That works just fine.

The constructor is declared as

A(std::fstream &file)

I'd bet the error you're keeping so super-secret is pretty informative toward answering your question (and I concur with the above, that it is an MVP). — WhozCraig, Feb 07 '14 at 00:37
@EdS: Seriously doubt this is MVP. `argv[1]` is not (ever) a type. — Lightness Races in Orbit, Feb 07 '14 at 00:47
@LightnessRacesinOrbit, [See this](http://coliru.stacked-crooked.com/a/9bb004535bb6526d). `a` is a function taking an array (pointer) of 1 `std::fstream` (the parameter name shadows `argv`) and returning an `A`. — chris, Feb 07 '14 at 01:03
@LightnessRacesinOrbit Thanks for this. It seems like this is a form of MVP, but I'm still not 100% clear. You said `a is a function taking an array (pointer)`. Is that correct? it seems like fstream is taking an array. Right? — markovuksanovic, Feb 07 '14 at 03:01
@chris: That looks like a bug, not MVP. The array parameter lacks a type. `argv[1]` is an _expression_. I'll look more into it. — Lightness Races in Orbit, Feb 07 '14 at 10:04
@chris: http://stackoverflow.com/q/21624880/560648 My bad :) — Lightness Races in Orbit, Feb 07 '14 at 10:12
@LightnessRacesinOrbit Many thanks for explaining this. This makes sense now, after reading the above mentioned question! — markovuksanovic, Feb 07 '14 at 11:28

score 6 · Accepted Answer · answered Feb 07 '14 at 00:42

You must have declared a() as follows:

A a(std::fstream &out);

This function a() takes a reference to a mutable (non-const) std::fstream. But, your code:

A a(fstream(argv[1]));

constructs a temporary std::fstream as part of the expression that is the function call to a(). C++ will not permit a temporary object to be used as an argument to a function taking a non-const reference.

The rationale for this rule is that the non-const-ness of the reference implies that the function wants to change the referred to object. Any changes to a temporary constructed in this way are "lost" in the sense that they are discarded when the temporary object is destroyed, as the function call expression finishes. So, this case is often a logical error, and the rule aims to "fail safe" and disallow it.

In your second case, the fin variable isn't a temporary, so there the rule doesn't apply.

+1: Not a bad introduction to Stack Overflow, whatsoever. Welcome! _(edit: bah; two years?!)_ — Lightness Races in Orbit, Feb 07 '14 at 00:48
Long time lurker ;) Decided I should answer some questions :) — Brad Spencer, Feb 07 '14 at 00:50

Why can't I pass function evaluation as constructor parameter in c++?

1 Answers1