Why do anonymous objects sometimes require a default constructor?

Question

If I write the following program, it works as I expect:

struct Foo {
    Foo (std::string x) { std::cout << x << std::endl; }
};

int main () { Foo("hello, world"); }

However, if I write a slightly different program, I get a compilation error:

struct Foo {
    Foo (std::string x) { std::cout << x << std::endl; }
};

std::string x("hello, world");

int main () { Foo(x); }

The error is:

prog.cc: In function 'int main()':
prog.cc:10:20: error: no matching function for call to 'Foo::Foo()'

The complete error can be seen on IDEONE.

Why does the error occur for the second program and not the first?

Moved this comment out of one answer, since it's apparently in _both_. This is another argument for C++11's [uniform initialization](http://www.stroustrup.com/C++11FAQ.html#uniform-init). — Drew Dormann, Jul 08 '14 at 01:54
@MooingDuck: Thanks, I forgot about the "redundant parentheses" rule for declarations. — jxh, Jul 08 '14 at 02:21

score 12 · Accepted Answer · answered Jul 08 '14 at 01:50

12

You have declared a variable x with type Foo

struct Foo {
    Foo(){}
    Foo (std::string x) { std::cout << x << std::endl; }
    void test(){ std::cout << "test" << std::endl; };
};

std::string x("hello, world");

int main () { Foo(x); x.test(); }

print "test"

What you want is use uniform initialization syntax Foo{x}

struct Foo {
    Foo (std::string x) { std::cout << x << std::endl; }
};

std::string x("hello, world");

int main () { Foo{x}; }

print "hello, world"

answered Jul 08 '14 at 01:50

Bryan Chen

45,816
18
112
143

1

Can you explain why `Foo a("a"), {x};` doesn't work? – jxh Jul 08 '14 at 02:09
2

@jxh what do you expect from it? `a` initialized with `"a"` and what? same reason you can have `int a(1), 2;`, you need to give a name for second object. – Bryan Chen Jul 08 '14 at 02:16
I am trying to create the anonymous temporary after the `,`. – jxh Jul 08 '14 at 02:18
@jxh it just not valid C++ syntax. – Bryan Chen Jul 08 '14 at 02:25
@jxh: How to do that would be an interesting SO question. – Mooing Duck Jul 08 '14 at 16:07
@MooingDuck: Except that it serves no real purpose, as I have no use case in mind. – jxh Jul 08 '14 at 16:51

score 8 · Answer 2 · edited May 23 '17 at 11:53

8

Foo(x); means the same as Foo x; . You can add redundant parentheses to the name of the variable.

So you need the default constructor because this is creating a variable x with no arguments being passed to constructor.

You can workaround this by doing Foo{x}; in C++11.

The parsing rule is that if something could be a valid declaration and a valid non-declaration statement, then it is in fact a declaration. See also most vexing parse

edited May 23 '17 at 11:53

Community

1
1

answered Jul 08 '14 at 01:49

M.M

138,810
21
208
365

"workaround"? No no no! It's the solution! It should be done for everything! Brace initialization is great! :-) – Mark Garcia Jul 08 '14 at 01:51
@MarkGarcia it seems to have its ups and downs... eg. `vector x{{"a", "b", "c"}};` is OK, but `vector x{{"a", "b"}};` causes undefined behaviour! – M.M Jul 08 '14 at 01:54
@Deduplicator, Not that I know of with the semicolon there. – chris Jul 08 '14 at 01:54
2

@MattMcNabb, And using single braces works for both :) I think brace-initialization is great when you know what to look out for. – chris Jul 08 '14 at 01:55
@chris: Please elaborate. What's wrong with a simple cast? – Deduplicator Jul 08 '14 at 01:56
@Deduplicator, As far as I know, `Type(identifier);` is always a declaration. `Type(identifier)` without the semicolon is a functional-style cast. Unfortunately, the plot thickens when `identifier` is replaced with `expression`. If you have a previous `x`, `Type(x);` is still a declaration. `Type(2);` is not. Even having a global `x` and doing `Type(::x);` fails to compile. – chris Jul 08 '14 at 01:57
1

What exactly is the problem with `Type(::x);` ? – M.M Jul 08 '14 at 02:05
1

@MattMcNabb, No clue. – chris Jul 08 '14 at 02:15
1

@MattMcNabb, It turns out `Type(::x);` compiles with GCC 4.9. I [asked a new question](http://stackoverflow.com/questions/24623071/is-typex-valid). – chris Jul 08 '14 at 03:08

score -1 · Answer 3 · answered Jul 08 '14 at 03:06

-1

I completely agree with above solution. Though this will still work:

struct Foo {
    Foo (string x) 
    { 
        cout << x << endl; 
    }
};


int main () 
{ 
    string x = "hello, world";
    Foo * abc = new Foo(x); 
    return 0;
}

answered Jul 08 '14 at 03:06

Abhishek Mittal

356
1
16

1

Not sure what this has to do with OP's question – P0W Jul 08 '14 at 03:18
I have just told a way by which we can still use the existing class definition of Foo. – Abhishek Mittal Jul 08 '14 at 03:22
Implicitly suggesting `using namespace std` and providing sample code with a memory leak, what an example! – Slava Feb 23 '17 at 20:34

Why do anonymous objects sometimes require a default constructor?

3 Answers3

Linked