constructor call or function-style cast in C++

Question

If I have the following c++ code:

class foo{
public:
    explicit foo(int i){};
};
void f(const foo &o){
}

And then I call

f(foo(1));

Is foo(1) constructor call or function-style cast?

Can't you only call constructors that way when declaring variables? — millimoose
– millimoose, Commented Nov 27, 2011 at 23:30
@Inerdial: not sure I understand you. Can you elaborate a bit? Thanks. — Qiang Li
– Qiang Li, Commented Nov 27, 2011 at 23:33
@QiangLi: I was asking if it's at all possible to call constructors without new as an expression, not as a variable declaration statement. (That said it's more of a vague guess which is why it's a comment not an answer.) — millimoose
– millimoose, Commented Nov 27, 2011 at 23:40
@Inerdial: it tends to get a bit vague when java/pythonists start guessing about C++ language features. Please post helpful comments when you're pretty certain you're contributing. — sehe
– sehe, Commented Nov 27, 2011 at 23:53

Oliver Charlesworth · Accepted Answer · 2011-11-27 23:30:47Z

4

They are the same thing.

answered Nov 27, 2011 at 23:30

Oliver Charlesworth

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5 Comments

Qiang Li Over a year ago

you mean I can consider them as either case?

sehe Over a year ago

LOL at  (I had to see how you managed that) +1

Qiang Li Over a year ago

@sehe: sorry about my English. What did you imply?

sehe Over a year ago

@QiangLi : that was at @OliCarlesworth :)

Steven Lu Over a year ago

(Edit the answer to see what it means... answers need to be padded to a certain length, so Oli used a HTML comment to do it. This is going in my bag of tricks)

Stuart Golodetz · Accepted Answer · 2011-11-27 23:31:52Z

4

It's a function-style cast that results in a constructor call, so both.

answered Nov 27, 2011 at 23:31

Stuart Golodetz

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1 Comment

Qiang Li Over a year ago

I thought only one is suffice. Then making a temporary copy to feed into the function is the next step thing. my question is only worrying about the code foo(1) part.

UncleBens · Accepted Answer · 2011-11-28 07:46:13Z

4

5.2.3 Explicit type conversion (functional notation)

1 A simple-type-specifier (7.1.6.2) or typename-specifier (14.6) followed by a parenthesized expression-list constructs a value of the specified type given the expression list. If the expression list is a single expression, the type conversion expression is equivalent (in definedness, and if defined in meaning) to the corresponding cast expression (5.4). ...

Your code creates a temporary, using the constructor you have with the argument's value 1, and binds it to a const reference. The temporary's lifetime ends at the end of the statement where it was created.

answered Nov 28, 2011 at 7:46

UncleBens

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