C++ FAQ / Section 10 / FAQ 10.21

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Revised Jul 4, 2012

[10.21] Why am I getting an error after declaring a Foo object via Foo x(Bar())?

Because that doesn't create a Foo object - it declares a non-member function that returns a Foo object.

This is really going to hurt; you might want to sit down.

First, here's a better explanation of the problem. Suppose there is a class called Bar that has a default ctor. This might even be a library class such as std::string, but for now we'll just call it Bar:

class Bar {
public:
  Bar();
  ...
};

Now suppose there's another class called Foo that has a ctor that takes a Bar. As before, this might be defined by someone other than you.

class Foo {
public:
  Foo(Bar const& b);  // or perhaps Foo(Bar b)
  ...
  void blah();
  ...
};

Now you want to create a Foo object using a temporary Bar. In other words, you want to create an object via Bar(), and pass that to the Foo ctor to create a local Foo object called x:

void yourCode()
{
  Foo x(Bar());  ← you think this creates a Foo object called x...
  x.blah();      ← ...but it doesn't, so this line gives you a bizarre error message
  ...
}

It's a long story, but one solution (hope you're sitting down!) is to add an extra pair of ()s around the Bar() part:

void yourCode()
{
  Foo x((Bar()));
        ^-----^----- ← These parens save the day
  x.blah();          ← Ahhhh, this works now — no more error messages
  ...
}

Another solution is to use = in your declaration (see the fine print below):

void yourCode()
{
  Foo x = Foo(Bar());  ← Yes, Virginia, that thar syntax works; see below for fine print
  x.blah();            ← Ahhhh, this works now — no more error messages
  ...
}

Note: The above solution requires yourCode() to be able to access the Foo copy constructor. In most situations that means the Foo copy constructor needs to be public, though it needn't be public in the less common case where yourCode() is a friend of class Foo. If you're not sure what any of that means, try it: if your code compiles, you passed the test.

Here's another solution (more fine print below):

void yourCode()
{
  Foo x = Bar();  ← Usually works; see below for fine print on "usually"
  x.blah();
  ...
}

Note: The word "usually" in the above means this: the above fails only when Foo::Foo(Bar const&) constructor is explicit, or when Foo's copy constructor is inaccessible (typically when it is private or protected, and your code is not a friend). If you're not sure what any of that means, take 60 seconds and compile it. You are guaranteed to find out whether it works or fails at compile-time, so if it compiles cleanly, it will work at runtime.

Okay, that's the end of the solutions; the rest of this is about why this is needed (this is optional; you can skip this section if you don't care enough about your career to actually understand what's going on; ha ha): When the compiler sees Foo x(Bar()), it thinks that the Bar() part is declaring a non-member function that returns a Bar object, so it thinks you are declaring the existence of a function called x that returns a Foo and that takes as a single parameter of type "non-member function that takes nothing and returns a Bar."

Now here's the sad part. In fact it's pathetic. Some mindless drone out there is going to skip that last paragraph, then they're going to impose a bizarre, incorrect, irrelevant, and just plain stupid coding standard that says something like, "Never create temporaries using a default constructor" or "Always use = in all initializations" or something else equally inane. If that's you, please fire yourself before you do any more damage. Those who don't understand the problem shouldn't tell others how to solve it. Harumph.

(Okay, that was mostly tongue in cheek. But there's a grain of truth in it. The real problem is that people tend to worship consistency, and they tend to extrapolate from the obscure to the common. That's not wise.)