assignment to type and empty braces. clarification on syntax

Question

I just came across some code and I believe I have never seen the syntax shown below.

struct A {
  int m_int;
  A (int a = int {}) : m_int(a) {}
};

So its clear that the constructor accepts an integer, by value, i.e. int a followed by assignment operator and type int, and empty braces & right parenthesis close. I am not able to decipher the latter part (int {}). What is meant by int a = int {}?. Please guide me to find out more about it. How do I address it?

Thank you, Gaurav

Apologies but I am referring to 'int a = int {}' part of the constructor syntax. — User9102d82, Jan 20 '20 at 19:15
https://stackoverflow.com/questions/35910619/default-argument-using-curly-braces-initializer — Mat, Jan 20 '20 at 19:16
https://stackoverflow.com/questions/33913572/brace-initialization-of-numeric-types-are-they-0-initialized — Mat, Jan 20 '20 at 19:19
a temporary is value initialized here 0 and that value is copied to the parameter. — Maestro, Jan 20 '20 at 19:20
The second `int` in `int a = int {}` is redundant. `int a = {}` does the same thing. — Jesper Juhl, Jan 20 '20 at 19:24
Thanks @Maestro. what benefit to we achieve by doing so. does this help us in any special way ? — User9102d82, Jan 20 '20 at 19:27
The new standard provides curly-braces initialization; one of the benefits is to prevent the narrowing conversion due to implicit conversions. — Maestro, Jan 20 '20 at 19:32
I am going to read more about the 'implicit conversions'. I was not aware of it. Need to study more. — User9102d82, Jan 20 '20 at 19:35

Maestro · Accepted Answer · 2020-01-20T19:35:32.003

2

In you constructor you are defining a constructor that takes an integer and the default constructor. Remember that a constructor that provides default arguments for all of its parameters defines also the default constructor.

You can write it this way:

A (int a = 0); // a is a default parameter.

In your example:

A (int a = int {});//

The parameter a is a default parameter initialized (not assigned) from a temporary integer which is value-initialized so because it is integer then it is value-initialized to 0 and then is used to initialize the parameter a. (a is a copy of it). Compiler optimize the code to remove the copy in many scenarios.

A (int a = int {5.6});// error
A (int a = int(5.6));// truncated to 5

edited Jan 20 '20 at 19:35

answered Jan 20 '20 at 19:29

Maestro

2,512
9
24

1

Yes. Now I understand it better. this was very helpful. Thanks again. – User9102d82 Jan 20 '20 at 19:34
You are welcome. – Maestro Jan 20 '20 at 19:37

score 1 · Answer 2 · answered Jan 20 '20 at 19:19

1

followed by assignment operator

No. This is not an assignment operation. This is syntax for default argument. It means that you can invoke the constructor without passing the argument explicitly, in which case int {} will be passed instead. The subexpression is syntax for value initialisation of a temporary object.

answered Jan 20 '20 at 19:19

eerorika

232,697
12
197
326

Yes. Rightly pointed out. It is not an assignment as I wrote it. – User9102d82 Jan 20 '20 at 19:22

assignment to type and empty braces. clarification on syntax

2 Answers2