• Brett (8/8) Sep 17 2019 T[N] t;
• Jonathan M Davis (10/18) Sep 17 2019 sort operates on random-access ranges. Static arrays are not ranges. In
• Jonathan Marler (4/12) Sep 17 2019 I believe T[N] t; sort(t); would pass t by value, so the sort
• Steven Schveighoffer (5/19) Sep 17 2019 std.algorithm.sort accepts a range. A fixed-sized array is not a range,

Brett <Brett gmail.com> writes:

T[N] t;
sort(t); fails

but

T[] t;
sort(t); passes


but magically

T[N] t;
sort(t[0..$]);  passes !!!

Jonathan M Davis <newsgroup.d jmdavisprog.com> writes:

On Tuesday, September 17, 2019 3:23:53 PM MDT Brett via Digitalmars-d wrote:
 T[N] t;
 sort(t); fails

 but

 T[] t;
 sort(t); passes


 but magically

 T[N] t;
 sort(t[0..$]);  passes !!!

sort operates on random-access ranges. Static arrays are not ranges. In
order for them to be ranges, it would have to be possible to pop elements
off of them, and that obviously won't work when the type has a fixed number
of elements. Slicing a static array results in a dynamic array which is a
slice of the static array, and that _is_ a random-access range, so it works
with sort.

BTW, if you want to slice the entirety of a static array, you don't need
indices. You can just do t[] rather than t[0..$].

- Jonathan M Davis

Jonathan Marler <johnnymarler gmail.com> writes:

On Tuesday, 17 September 2019 at 21:23:53 UTC, Brett wrote:
 T[N] t;
 sort(t); fails

 but

 T[] t;
 sort(t); passes


 but magically

 T[N] t;
 sort(t[0..$]);  passes !!!

I believe T[N] t; sort(t); would pass t by value, so the sort 
function would actually do nothing to t, it would only operate on 
a copy of t that was passed on the stack.

Steven Schveighoffer <schveiguy gmail.com> writes:

On 9/17/19 5:23 PM, Brett wrote:
 T[N] t;
 sort(t); fails
 
 but
 
 T[] t;
 sort(t); passes
 
 
 but magically
 
 T[N] t;
 sort(t[0..$]);  passes !!!
 

std.algorithm.sort accepts a range. A fixed-sized array is not a range, 
since you can't pop the front part off and make it 1 smaller.

You can use sort(t[]) for the same effect.

-Steve