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Getting c++11 auto initialization syntax right


Getting c++11 auto initialization syntax right

By : Blaise Vignon
Date : November 22 2020, 01:01 AM
hop of those help? Since all the answers were in the comments, I thought I'd finish this off by doing an official answer myself.
I am using a c++ library that doesn't have movable streams, and this matters because
code :
auto is2 = std::istream{&cmdStreamBuf_};
auto varname = typename{...};
typename varname{...};


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C++ : 3 questions about initialization syntax, value-initialization and default-initialization

C++ : 3 questions about initialization syntax, value-initialization and default-initialization


By : Himanshu Dwivedi
Date : March 29 2020, 07:55 AM
it fixes the issue
The notation int x(1); is called direct-initialization. It's defined separately for native types and for classes. In the latter case a constructor would be called.
In-class member initialization with an initializer list using uniform initialization syntax?

In-class member initialization with an initializer list using uniform initialization syntax?


By : Aditya Srivastava
Date : March 29 2020, 07:55 AM
I wish did fix the issue. I am trying to compile the following with MSVC2013: , Here is the responsible grammar from N3797:
code :
// after a member declaration:
braced-or-equal-initializer-list:
  = initializer-clause
  braced-init-list

braced-init-list:
  { initializer-list ,OPT }
  { }

initializer-list:
  initializer-clause
  initializer-list, initializer-clause

initializer-clause:
  assignment-expression
  braced-init-list
Confusion with auto assignment with list initialization syntax

Confusion with auto assignment with list initialization syntax


By : Andrew Gillespie
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , When you use the functional notation for type conversion, the type name must be a simple-type-specifier or typename-specifier (§5.2.3 [expr.type.conv]), which basically means a type name consisting of a single word.
code :
unsigned int a = unsigned int(42);
foo const* f{nullptr};
using foo_const_p = foo const*;
auto f = foo_const_p{nullptr};
Difference between (auto) properties initialization syntax in C# 6

Difference between (auto) properties initialization syntax in C# 6


By : AshokB
Date : March 29 2020, 07:55 AM
wish helps you Listing 3 is C# 6's equivalent of listing 2, where the backing field is provided under the hood.
Listing 4:
code :
public List<string> Items => new List<string>();
public List<string> Items { get { return new List<string>(); } }
`auto x = type{...}` initialization syntax and `explicit` conversion operator - clang vs gcc

`auto x = type{...}` initialization syntax and `explicit` conversion operator - clang vs gcc


By : Abhi
Date : March 29 2020, 07:55 AM
Hope this helps Using http://open-std.org/JTC1/SC22/WG21/docs/papers/2016/n4606.pdf I think g++ is wrong.
8.6.4 clause 3.7 states:
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