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digitalmars.D.bugs - [Issue 12173] New: Optional start value for std.algorithm.sum

reply d-bugmail puremagic.com writes:
https://d.puremagic.com/issues/show_bug.cgi?id=12173

           Summary: Optional start value for std.algorithm.sum
           Product: D
           Version: D2
          Platform: x86
        OS/Version: Windows
            Status: NEW
          Severity: enhancement
          Priority: P2
         Component: Phobos
        AssignedTo: nobody puremagic.com
        ReportedBy: bearophile_hugs eml.cc



The built-in Python sum() function supports a second optional argument that is
the start, it's useful when you want to sum an iterable of values starting from
a seed of another type (of different from zero):


 sum([1, 2, 3])






6

 sum([1, 2, 3], 2)






8

 sum([1, 2, 3], 0.0)






6.0

 sum([[1, 2], [3]], [])






[1, 2, 3]


This shows one use case for the same functionality in D:


import std.algorithm: sum, reduce;
import std.functional: curry;
struct Foo {
    ubyte x;
    alias x this;
}
alias mySum = curry!(reduce!q{a + b}, 0);
void main() {
    Foo[] arr;
    arr.mySum;    // OK
    arr.sum;      // Error
    //arr.sum(0); // Not supported
}


dmd 2.065beta3 gives:

..\dmd2\src\phobos\std\algorithm.d(1087,19): Error: cannot implicitly convert
expression (0) of type int to Foo

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next sibling parent d-bugmail puremagic.com writes:
https://d.puremagic.com/issues/show_bug.cgi?id=12173


Andrei Alexandrescu <andrei erdani.com> changed:

           What    |Removed                     |Added
----------------------------------------------------------------------------
                 CC|                            |andrei erdani.com



PST ---
Doesn't an add outside the sum call do the same?

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prev sibling parent d-bugmail puremagic.com writes:
https://d.puremagic.com/issues/show_bug.cgi?id=12173






 Doesn't an add outside the sum call do the same?


An add outside the sum doesn't solve the problem with Foo.

Another example:

long[] data = [...];
immutable total = data.sum(0.BigInt);

If the longs are large enough this gives the correct result, unlike summing
with longs.

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