• Piotr Szturmaj (5/5) Aug 23 2012 Hi,
• Timon Gehr (4/9) Aug 23 2012 It is a method of a struct, therefore it is not parameterless, but has
• Piotr Szturmaj (14/26) Aug 23 2012 Thank you. So, it's helpful because struct might be qualified somewhere
• Kagamin (2/2) Aug 24 2012 Well, it may be also a way to ensure the function doesn't modify

Piotr Szturmaj <bncrbme jadamspam.pl> writes:

Hi,

I found this code of std.range.iota's Result struct:

          property inout(Value) front() inout { assert(!empty); return 
current; }

What's the purpose of inout on parameterless functions?

Timon Gehr <timon.gehr gmx.ch> writes:

On 08/24/2012 12:14 AM, Piotr Szturmaj wrote:
 Hi,

 I found this code of std.range.iota's Result struct:

           property inout(Value) front() inout { assert(!empty); return
 current; }

 What's the purpose of inout on parameterless functions?

It is a method of a struct, therefore it is not parameterless, but has
a hidden parameter. 'inout' qualifies the implicit 'this' reference.
Inside the method body, typeof(this) is inout(Result).

Piotr Szturmaj <bncrbme jadamspam.pl> writes:

Timon Gehr wrote:
 On 08/24/2012 12:14 AM, Piotr Szturmaj wrote:
 Hi,

 I found this code of std.range.iota's Result struct:

           property inout(Value) front() inout { assert(!empty); return
 current; }

 What's the purpose of inout on parameterless functions?

 It is a method of a struct, therefore it is not parameterless, but has
 a hidden parameter. 'inout' qualifies the implicit 'this' reference.
 Inside the method body, typeof(this) is inout(Result).

Thank you. So, it's helpful because struct might be qualified somewhere 
as const or as immutable. Anyway in that particular case it's 
unnecessary because iota's Result must be mutable to call popFront().

     immutable iotaRange = iota(0, 5);

     pragma(msg, typeof(&iotaRange.front));
     pragma(msg, typeof(iotaRange.front));

     io.popFront();

yields:

     inout(int) delegate() inout  property
     immutable(int)
     main.d(61): Error: function 
std.range.iota!(int,int).iota.Result.popFront () is not callable using 
argument types () immutable

"Kagamin" <spam here.lot> writes:

Well, it may be also a way to ensure the function doesn't modify 
the struct.