How does std::string construct an object with the name as its constructor argument?

Question

#include <string>          
#include <iostream>        

std::string(foo);          

int main()                 
{                          
    std::cout << foo.size() << "\n";                                                
    return 0;              
}

Results in 0, instead of an expected compile error for foo being undefined.

How is it able to do this? What is this called?

If doesn't. That's equivalent to `std::string foo;` You declared an object foo of type `std:string`. — Panagiotis Kanavos, Feb 19 '19 at 10:55

P.W · Answer 1 · 2019-02-19T11:07:37.767

11

std::string(foo);  //#1

is the same as

std::string (foo); //#2

is the same as

std::string foo; //#3

The parentheses in #2 are redundant. They are needed in #1 as there is no whitespace separating std::string and foo.

edited Feb 19 '19 at 11:07

answered Feb 19 '19 at 10:57

P.W

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Great answer, much easier to understand than the linked duplicate. – Mark Ransom Feb 19 '19 at 17:27
@MarkRansom: Thank you. But it is hard to convince the "hammer-holders" about this. :-) – P.W Feb 20 '19 at 04:41

score 0 · Answer 2 · answered Feb 19 '19 at 10:57

The definition std::string(foo); calls the default constructor of std::string. The default constructor by definition is the one that can be called without arguments, and indeed you do not provide one.

A default-constructed string is empty, so its size is indeed 0.

How does std::string construct an object with the name as its constructor argument?

2 Answers2