The below method works flawlessly
public <T> void fromJsonArray(String jsonString,Type tToken) {
Gson g = new Gson();
T list = g.fromJson(jsonString,tToken);
System.out.println(list);
}
But i have not specified what < T > is in this method. how does the compiler assign the value returned by the fromJson
method to the variable list
whose type i have not specified?
i just tested the validity of the answers stating about <T>
being inferred from the return type of the method. It doesn't seem to work out. Please check out the following code. it doesn't even compile
import java.util.*;
class Sample {
public List<String> getT(String s) {
List<String> list = new ArrayList<String>();
list.add(s);
return list;
}
public <T> void test(){
T list = getT("test");
System.out.println(l);
}
public static void main(String[] a) {
new Sample().test();
}
}
modified the source again and tested it, and it resulted in a compile time error
public <T> List<T> getT(T s) {
List<T> list = new ArrayList<T>();
list.add(s);
return list;
}
public <T> void test(){
T list = getT("test"); //incompatible types compilation error here
System.out.println(list);
}
Sample1.java:13: error: incompatible types
T list = getT("test");
^
required: T
found: List
where T is a type-variable:
T extends Object declared in method test()
Some languages require every symbol to be explicitly typed. Python allows a symbol to be bound to different values at runtime, so its type can change over time. A symbol's type doesn't need to be declared statically.
The <T> parameter tells the compiler that both List s being passed as well as the List returned must be of the same generic type.
Type inference is a Java compiler's ability to look at each method invocation and corresponding declaration to determine the type argument (or arguments) that make the invocation applicable.
Type inference represents the Java compiler's ability to look at a method invocation and its corresponding declaration to check and determine the type argument(s). The inference algorithm checks the types of the arguments and, if available, assigned type is returned.
how does the method infer the type of
<T>
It doesn't. Generic methods don't infer their generic types - that's why T
is called a type parameter. The caller of the method provides a type argument for T
. When it does, it may be inferred by the compiler based on the context of the method call's arguments and target type.
For example:
Set<String> c = Collections.emptySet();
emptySet
declares a type parameter T
, takes no arguments, and returns a Set<T>
. Here, the compiler infers T
to be String
based on the target type, Set<String>
.
Another example:
Collections.singleton("asdf");
singleton
declares a type parameter T
, takes a T
, and returns a Set<T>
. Here, there is no target type, but the compiler infers T
to be String
based on the argument "asdf"
.
But generic type inference is just a convenience. Without it, we could still use type witnesses to explicitly provide type arguments:
Set<String> c = Collections.<String>emptySet();
Collections.<String>singleton("asdf");
This brings us to your method signature:
public <T> void fromJsonArray(String jsonString, Type tToken)
fromJsonArray
declares a type parameter T
, but doesn't return anything related to the type T
or take arguments related to T
. At a call to fromJsonArray
, the compiler has no information from which to infer T
. Its type argument will default to its upper bound Object
unless a type witness is used:
myObj.<String>fromJsonArray(jsonString, tToken);
But this doesn't matter because <String>
has no affect on the behavior of the method call or its compilation. T
is meaningless* and can be removed from the declaration of fromJsonArray
.
how does the compiler assign the value returned by the
fromJson
method to the variablelist
whose type i have not specified?
Here is the source of Gson.fromJson(String, Type)
:
@SuppressWarnings("unchecked")
public <T> T fromJson(String json, Type typeOfT) throws JsonParseException {
StringReader reader = new StringReader(json);
T target = (T) fromJson(reader, typeOfT);
return target;
}
You can see it declares an arbitrary type parameter T
and casts the deserialized object to T
. This is known as an unchecked cast, because it won't fail fast if it's wrong. That's because T
has been erased at runtime. You can see that the code is suppressing a warning about doing this, because it's generally a bad idea. By not restricting what T
is based on the method arguments, the Gson code has effectively ceded control over it to the caller. If you wrote:
List<String> list = g.fromJson(jsonString, tToken);
but tToken
represented HashSet<String>
, you would get a ClassCastException
on that line at runtime. Worse, if tToken
represented ArrayList<Integer>
, it would not even fail on that line, because the JVM would only see a List
and allow the assignment to happen. A ClassCastException
would be thrown sometime later, once your code tried to treat the list's Integer
elements like String
s (and the exception would be confusing to debug).
So to answer your question about the assignment, the compiler lets you assign the result of fromJson
to anything you want. It's up to you for it to be correct.
You may ask, Why would Gson do an unchecked cast and allow unsafe code? The answer is that it's a convenience, stemming from language limitations. Their other signature is safer:
public <T> T fromJson(String json, Class<T> classOfT)
But there's no way to represent generic types with Class
- no List<String>.class
for example. Only a Type
can do this, and it's not itself generic. fromJson
could have required a TypeToken<T>
, but there are other ways to obtain a Type
, so that would be restrictive.
Returning Object
and forcing the caller to do the unchecked cast would have been more transparent but the Gson developers probably wanted to avoid this "ugliness".
T list = g.fromJson(jsonString,tToken);
It's inferred from the return type of g.fromJson().
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