Submitted *Issue 4883 <http://d.puremagic.com/issues/show_bug.cgi?id=4883>
*
On Fri, Sep 17, 2010 at 4:30 PM, Steven Schveighoffer
<schvei...@yahoo.com>wrote:
> On Fri, 17 Sep 2010 17:15:31 -0400, Seth Hoenig <seth.a.hoe...@gmail.com>
> wrote:
>
> I have these two minimal programs:
>>
>>
>> import std.string;
>> void main()
>> {
>> string str = "abc";
>> int i = str.count("ab");
>>
>> }
>>
>>
>>
>> and:
>>
>>
>>
>> import std.string;
>> import std.algorithm;
>> void main()
>> {
>> string str = "abc";
>> int i = str.count("ab");
>>
>> }
>>
>>
>>
>> The only difference is line 2, where I import std.algorithm.
>> The first program compiles fine, but the second program does not compile,
>> spitting out the error message:
>>
>> bash-3.2$ dmd -ofdummy dummy.d
>> /u/sah2659/dmd2/linux/bin/../../src/phobos/std/functional.d(176): Error:
>> static assert "Bad binary function q{a == b}. You need to use a valid D
>> expression using symbols a of type dchar and b of type string."
>> /u/sah2659/dmd2/linux/bin/../../src/phobos/std/functional.d(179):
>> instantiated from here: Body!(dchar,string)
>> /u/sah2659/dmd2/linux/bin/../../src/phobos/std/algorithm.d(3410):
>> instantiated from here: result!(dchar,string)
>> dummy.d(7): instantiated from here: count!("a == b",string,string)
>>
>>
>> I can't imagine I'm the first person to notice a bug like this, so is
>> there
>> something I am doing wrong?
>>
>
> I see two bugs here. First, this should be an ambiguity error, because
> count matches both std.algorithm.count and std.string.count. The compiler
> should refuse to compile this I think.
>
> Second, std.algorithm.count looks like this:
>
> size_t count(alias pred = "a == b", Range, E)(Range r, E value) if
> (isInputRange!(Range))
>
> So, E can be any type, completely unrelated to strings, I could do:
>
> count!(string, int[]) which makes no sense.
>
> I think the sig should be
>
> size_t count(alias pred = "a == b", Range, E)(Range r, E value) if
> (isInputRange!(Range) && isImplicitlyConvertable!(E, ElementType!(Range)))
>
> -Steve
>
