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For Lambda expressions, the compiler doesn't translate them into something which is already understood by JVM. Lambda syntax that is written by the developer is desugared into JVM level instructions generated during compilation, which means the actual responsibility of constructing lambda is deferred to runtime.
The value of a lambda expression is a reference to an instance of a class with the following properties: The class implements the targeted functional interface and, if the target type is an intersection type, every other interface element of the intersection.
A return statement is not an expression in a lambda expression. We must enclose statements in braces ({}). However, we do not have to enclose a void method invocation in braces. The return type of a method in which lambda expression used in a return statement must be a functional interface.
Lambda functions are pure because they do not rely on a specific class scope. They are immutable because they reference the passed parameter but do not modify the parameter's value to reach their result. Finally, they're first-class functions because they can be anonymous and passed to other functions.
As I know all data types must be known at compile time, and lambda is not a type. Does lambda got translated into anonymous struct with operator() or std::function wrapped?
For example,
std::for_each(v.begin(), v.end(), [](int n&){n++;});
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