Re: importing std.algorithm breaks std.string.count

[Steven Schveighoffer]

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