Java Generics, Type Inference, Inheritance? -

i'm reading type inference generics, , code provided example fails compile.

import java.io.*; class lasterror<t> {     private t lasterror;      void seterror(t t){         lasterror = t;         system.out.println("lasterror: seterror");     } }  class strlasterror<s extends charsequence> extends lasterror<string>{     public strlasterror(s s) {     }     void seterror(s s){         system.out.println("strlasterror: seterror");     } } class test {     public static void main(string []args) {         strlasterror<string> err = new strlasterror<string>("error");         err.seterror("last error");     } } 

and explanation given in book was:

"(it looks seterror() method in strlasterror overriding seterror() in lasterror class. however, not case. @ time of compilation, knowledge of type s not available. therefore, compiler records signatures of these 2 methods seterror(string) in superclass , seterror(s_extends_charsequence) in subclass—treating them overloaded methods (not overridden). in case, when call seterror() found, compiler finds both overloaded methods matching, resulting in ambiguous method call error."

i don't understand why type s can't inferred @ compile time. string passed when invoking constructor of class strlasterror, , api docs, string implement interface charsequence, doesn't mean s <s extends charsequence> of type string?

i've read java online tutorial on topic of generics several times. i've checked "type inference", , inheritance, don't know how whole thing works. need explanation on question.

the points i'm stuck @ are:

1. if subtype can't decide s, how come super type can decide t, because superclass not have upper bound? or infer t string because subtype calls supertype's constructor first?

2. i understand if constructor invoked as:

   strlasterror<charsequence> err = newstrlasterror<>((charsequence)"error"); 

   there no ambiguity, since it's plain method overriding then. (or wrong here?)

however, said in beginning, if string passed, why can't s inferred string?

if subtype can't decide s, how come super type can decide t, because superclass not have upper bound? or infer t string because subtype calls supertype's constructor first?

the super type isn't deciding or inferring t is; you're explicitly telling t declaration:

class strlasterror<s extends charsequence> extends lasterror<string> 

t bound string lasterror, makes every reference t in parent class concrete string.

your child class has bound s extends charsequence attached it, independent bounds applied parent class.

what happens java compile child class, , result of child class 2 methods signature matches string created. (key note here: a string is-a charsequence.)

in child class, seterror(ljava/lang/charsequence;)v generated signature seterror. because of way generics work, lasterror#seterror treated if has signature of seterror(ljava/lang/string;)v. why when go override method, place string type parameter instead of else.

so, arrive @ two methods have override-equivalent signatures.

void seterror(charsequence s) void seterror(string s) 

jls 8.4.8.4. applies here.

it possible class inherit multiple methods override-equivalent signatures (§8.4.2).

it compile-time error if class c inherits concrete method signature subsignature of concrete method inherited c. can happen if superclass generic, , has 2 methods distinct in generic declaration, have same signature in particular invocation used.

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i understand if constructor invoked strlasterror err = new strlasterror<>((charsequence)"error"); there no ambiguity, since plain method overriding then.(or i'm wrong here)

no, you're messing raw types. interestingly enough work, because signatures 2 methods has become:

void seterror(object s) void seterror(string s) 

you want use generics avoid scenario this; may want invoke super class method @ point, in scenario these bindings, it's difficult accomplish.