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digitalmars.D.learn - A matter of inout

reply "bearophile" <bearophileHUGS lycos.com> writes:
This program used to compile in DMD 2.059:


import std.stdio;
inout(T[]) forwardDifference(T)(inout(T[]) s, in int n) pure {
     foreach (_; 0 .. n)
         s[] -= s[1 .. $];
     return s[0 .. $-n];
}
void main() {
     immutable A = [90.5, 47, 58, 29, 22, 32, 55, 5, 55, 73.5];
     foreach (level; 0 .. A.length)
         writeln(forwardDifference(A.dup, level));
}


Now it gives:

test.d(5): Error: slice s[] is not mutable
test.d(12): Error: template instance 
test.forwardDifference!(double) error instantiating

Is it a 2.060 regression (or it's caused by a bug fix)?

Bye,
bearophile
Jul 27 2012
next sibling parent Dmitry Olshansky <dmitry.olsh gmail.com> writes:
On 27-Jul-12 19:53, bearophile wrote:

 This program used to compile in DMD 2.059:


 import std.stdio;
 inout(T[]) forwardDifference(T)(inout(T[]) s, in int n) pure {
      foreach (_; 0 .. n)
          s[] -= s[1 .. $];
      return s[0 .. $-n];
 }
 void main() {
      immutable A = [90.5, 47, 58, 29, 22, 32, 55, 5, 55, 73.5];
      foreach (level; 0 .. A.length)
          writeln(forwardDifference(A.dup, level));
 }


 Now it gives:

 test.d(5): Error: slice s[] is not mutable
 test.d(12): Error: template instance test.forwardDifference!(double)
 error instantiating

 Is it a 2.060 regression (or it's caused by a bug fix)?


If it compiled before then I'd think it's a regression.



-- 
Dmitry Olshansky
Jul 27 2012
prev sibling parent reply Artur Skawina <art.08.09 gmail.com> writes:
On 07/27/12 17:53, bearophile wrote:

 This program used to compile in DMD 2.059:
 
 
 import std.stdio;
 inout(T[]) forwardDifference(T)(inout(T[]) s, in int n) pure {
     foreach (_; 0 .. n)
         s[] -= s[1 .. $];
     return s[0 .. $-n];
 }
 void main() {
     immutable A = [90.5, 47, 58, 29, 22, 32, 55, 5, 55, 73.5];
     foreach (level; 0 .. A.length)
         writeln(forwardDifference(A.dup, level));
 }
 
 
 Now it gives:
 
 test.d(5): Error: slice s[] is not mutable
 test.d(12): Error: template instance test.forwardDifference!(double) error
instantiating
 
 Is it a 2.060 regression (or it's caused by a bug fix)?


'inout' basically means 'local const' - if you'd allow inout
data to be modified then it wouldn't be possible to call the
function with a const or immutable argument - which is the
whole point of inout...
You probably want 'inout(T)[]' if you're going to alter the
array.

artur
Jul 27 2012
parent reply "bearophile" <bearophileHUGS lycos.com> writes:
Artur Skawina:


 'inout' basically means 'local const' - if you'd allow inout
 data to be modified then it wouldn't be possible to call the
 function with a const or immutable argument - which is the
 whole point of inout...
 You probably want 'inout(T)[]' if you're going to alter the
 array.


Using inout(T)[] gives the same error:

import std.stdio;
inout(T)[] forwardDifference(T)(inout(T)[] s, in int n) pure {
     foreach (_; 0 .. n)
         s[] -= s[1 .. $];
     return s[0 .. $ - n];
}
void main() {
     immutable A = [90.5, 47, 58, 29, 22, 32, 55, 5, 55, 73.5];
     foreach (level; 0 .. A.length)
         writeln(forwardDifference(A.dup, level));
}


Bye,
bearophile
Jul 27 2012
next sibling parent reply "bearophile" <bearophileHUGS lycos.com> writes:
Artur Skawina:


 'inout' basically means 'local const' - if you'd allow inout
 data to be modified then it wouldn't be possible to call the
 function with a const or immutable argument - which is the
 whole point of inout...


dmd 2.059 was wrong then.

Bye,
bearophile
Jul 27 2012
parent =?UTF-8?B?QWxpIMOHZWhyZWxp?= <acehreli yahoo.com> writes:
On 07/27/2012 10:13 AM, bearophile wrote:

 Artur Skawina:


 'inout' basically means 'local const' - if you'd allow inout
 data to be modified then it wouldn't be possible to call the
 function with a const or immutable argument - which is the
 whole point of inout...


 dmd 2.059 was wrong then.


I think so.

I have understood that aspect of inout recently: since the function is 
not a template (bear with me please! I know it is a template here :)), 
there will be a single instance of it. Since that instance must work 
with mutable and immutable, the parameter cannot be modified inside the 
function.

Even if the function is never called with immutable, the compiler must 
compile the code as if 'inout' is spelled as 'const'.

The above makes sense for a regular function. I guess the guideline here 
is not to use inout on a template parameter as it doesn't make sense 
because T carries that information anyway.

Ali
Jul 27 2012
prev sibling parent Artur Skawina <art.08.09 gmail.com> writes:
On 07/27/12 19:06, bearophile wrote:

 Artur Skawina:
 

 'inout' basically means 'local const' - if you'd allow inout
 data to be modified then it wouldn't be possible to call the
 function with a const or immutable argument - which is the
 whole point of inout...
 You probably want 'inout(T)[]' if you're going to alter the
 array.


 
 Using inout(T)[] gives the same error:
 
 import std.stdio;
 inout(T)[] forwardDifference(T)(inout(T)[] s, in int n) pure {
     foreach (_; 0 .. n)
         s[] -= s[1 .. $];
     return s[0 .. $ - n];
 }
 void main() {
     immutable A = [90.5, 47, 58, 29, 22, 32, 55, 5, 55, 73.5];
     foreach (level; 0 .. A.length)
         writeln(forwardDifference(A.dup, level));
 }


This alters not only the array, but also modifies the elements.

Using just 'T' instead of 'inout(T)' will do the right thing;
why would you like to use inout in cases like this one? 'inout'
allows you to not write two or more /identical/ function bodies
when the code can deal with immutable/const/mutable; functions
that modify the data in-place obviously can't do that.

artur
Jul 27 2012