What do parameterless constructors mean when explicitly called at declaration

Question

The question sounds a bit odd, but check the code sample

#include <iostream>

using namespace std;

int main() {
    string a, b(), c("test");

    // No problems
    a = c;

    // tester.cpp:9:7: error: assignment of function ‘std::string b()’
    b = c;

    // tester.cpp:10:7: error: invalid conversion from ‘std::string (*)() {aka std::basic_string<char> (*)()}’ to ‘char’
    a = b;

    // No problems
    c = a;
}

You can see that it looks like b is created with the default constructor, but in fact it is not. So my question is really is, what does string b() mean?

Indeed almost a dupe, but I think worth answering due to the commas adding to the melee. — Bathsheba, Sep 19 '14 at 08:25

Bathsheba · Accepted Answer · 2014-09-19T08:25:15.460

1

Condensing your code to string b();, what you're doing here is declaring a prototype to a function called b that returns a string and takes no arguments.

The commas in the "compound declaration" are obfuscating this.

edited Sep 19 '14 at 08:25

answered Sep 19 '14 at 07:59

Bathsheba

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Isn't calling it the "comma operator" a little misleading in itself? It's a list of declarators separated by commas, the comma operator is a different beast. – user657267 Sep 19 '14 at 08:15
I think my terminology is correct; follows the C++ grammar. – Bathsheba Sep 19 '14 at 08:18
@Bathsheba: Operators appear in expressions, not in declarator lists. This is indeed a declarator list. – MSalters Sep 19 '14 at 08:22
e.g. `int i = 1, 2;` won't compile, the comma here is not an operator. – user657267 Sep 19 '14 at 08:23
OK. I see your point. I've amended. – Bathsheba Sep 19 '14 at 08:25

score 1 · Answer 2 · answered Sep 19 '14 at 07:59

1

It's actually a function declaration. It's possible to declare local functions but not to define them. So therefore it's not possible to explicitly call the default constructor.

answered Sep 19 '14 at 07:59

Fytch

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score 1 · Answer 3 · answered Sep 19 '14 at 07:59

1

Classic mistake, but the error message already says it: function std::string b() /.

answered Sep 19 '14 at 07:59

MSalters

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What do parameterless constructors mean when explicitly called at declaration

3 Answers3