java - 这个 "s"实际上是最终的吗？

Question

这是一些示例代码(假设是 Java 8)。

while (true){
    Socket s = serverSocket.accept();
    // some code here ... we don't assign anything to s again here ... 
}

s 在循环中是有效的 final 吗？

Answer 1

假设 s 没有在其声明之后被赋值，那么是的，根据 JLS, Section 4.12.4，s 是“有效最终的” ，它定义了它，在你的情况下:

Certain variables that are not declared final are instead considered effectively final:

• A local variable whose declarator has an initializer (§14.4.2) is effectively final if all of the following are true:

  • It is not declared final.

  • It never occurs as the left hand side in an assignment expression (§15.26). (Note that the local variable declarator containing the initializer is not an assignment expression.)

  • It never occurs as the operand of a prefix or postfix increment or decrement operator (§15.14, §15.15).

• A local variable whose declarator lacks an initializer is effectively final if all of the following are true:

  • It is not declared final.

  • Whenever it occurs as the left hand side in an assignment expression, it is definitely unassigned and not definitely assigned before the assignment; that is, it is definitely unassigned and not definitely assigned after the right hand side of the assignment expression (§16 (Definite Assignment)).

  • It never occurs as the operand of a prefix or postfix increment or decrement operator.

您只能在声明时分配 s，作为一个对象，它不能作为递增或递减运算符的操作数，因此它实际上是最终的。

它还说明了一个变量在另一种情况下可以是有效的final，如果它在声明时没有被赋值，只要它只被赋值一次，它肯定在声明之前没有被赋值，它肯定在声明之后被赋值。

此外，在该部分的末尾，它指出:

If a variable is effectively final, adding the final modifier to its declaration will not introduce any compile-time errors. Conversely, a local variable or parameter that is declared final in a valid program becomes effectively final if the final modifier is removed.

您应该能够显式地使其成为 final 而不会导致编译器错误。如果是这样，它实际上是最终的。