c# - Why doesn't/couldn't IDictionary<TKey,TValue> implement ILookup<TKey,TValue>? -

-

i suppose doesn't matter, i'm curious.

if difference between dictionary , lookup 1 one-to-one , other one-to-many, wouldn't dictionary more specific/derived version of other?

a lookup collection of key/value pairs key can repeated. dictionary collection of key/value pairs key cannot repeated.

why couldn't idictionary implement ilookup?

i suspect because intention different.

ilookup<t,u> designed work collection of values. idictionary<t,u> intended work single value (that could, of course, collection).

while could, of course, have idictionary<t,u> implementations implement via returning ienumerable<u> single value, confusing, if "u" collection (ie: list<int>). in case, ilookup<t,u>.item return ienumerable<list<int>>, or should type of check ienumerable<t> value type, , "flatten" it? either way, it'd confusing, , add questionable value.


Comments

Post a Comment

Popular posts from this blog

c# - Is it possible to remove an existing registration from Autofac container builder? -

-
something along lines: builder.registertype<mytype>().as<itype>(); builder.registertype<mytype2>().as<itype>(); builder.deregistertype<mytype>().as<itype>() var container = builder.build(); var types = container.resolve<ienumerable<itype>>(); assert.istrue(types.count == 1); assert.istrue(types[0].gettype == typeof(mytype2)); scenario: go through bunch of assemblies , go register types want make sure have 1 implementation of given type. need before create container. track on own nice if autofac me bit. this cannot done directly using containerbuilder , unless start on new one. mind you, having first built container should able construct new container filtering away unwanted types , reusing registrations first container. this: ... var container = builder.build(); builder = new containerbuilder(); var components = container.componentregistry.registrations .where(cr => cr.activator.limittype != typ...
Read more

c++ - Warning : overflow in implicit constant conversion -

-
in following program, line 5 give overflow warning expected, surprisingly line 4 doesn't give warning in gcc: http://www.ideone.com/u0bxn int main() { int = 256; char c1 = i; //line 4 char c2 = 256; //line 5 return 0; } i thinking both lines should give overflow warning. or there i'm missing? the topic led me experiment this: typedef type checking? there said following(which deleted answer, because when run it, didn't show had expected): //however, you'll warning case: typedef int t1; typedef char t2; t1 x = 256; t2 y = x; //possible overflow warning! (but doesn't give warning :() -wall doesn't include many options. -wconversion 1 of them , warns behavior you're interested in. see http://gcc.gnu.org/onlinedocs/gcc/warning-options.html
Read more