在循环中为局部变量分配空间 [英] Allocation of space for local variables in loops

问题描述

(是或否)在执行循环主体时分配在循环主体中声明的局部变量的空间，并在主体结束时释放.

(true or false) The space for a local variable that is declared in the body of the loop is allocated whenever the loop body is executed and deallocated when the body finishes.

这个问题的答案是错误的.但是为什么呢?

The answer to this question is false. But why?

推荐答案

该语句为false，因为未分配和释放局部变量空间.它存在于堆栈中，并在输入方法时保留.

The statement is false because local variable space is not allocated and deallocated. It exists on the stack and is reserved when the method is entered.

要查看如何使用堆栈空间，请使用以下代码编写一个小型测试程序:

To see how stack space is used, write a small test program with:

public static void test() {
    {
        int a = 1;
        long b = 2;
        int c = 3;
    }
    {
        int x = 4;
        int y = 5;
        long z = 6;
    }
}

现在，使用以下命令将其反汇编以查看字节码.

Now disassemble it with the following command to see the bytecode.

javap -c Test.class

这是输出.为了方便起见，我在右侧添加了Java代码.

Here is the output. I've added the Java code on the right for your convenience.

  public static void test();
    Code:
       0: iconst_1                     int a = 1;
       1: istore_0
       2: ldc2_w     #22   long 2l     long b = 2;
       5: lstore_1
       6: iconst_3                     int c = 3;
       7: istore_3
       8: iconst_4                     int x = 4;
       9: istore_0
      10: iconst_5                     int y = 5;
      11: istore_1
      12: ldc2_w     #24   long 6l     long z = 6;
      15: lstore_2
      16: return                       return;

发生的事情是该方法保留了4个插槽". int 变量占用1个插槽，而 long 变量占用2个插槽.

What happens is that the method has reserved 4 "slots". An int variable takes 1 slot, and a long variable takes 2 slots.

所以代码确实说:

slot[0] = 1
slot[1-2] = 2L
slot[3] = 3

slot[0] = 4
slot[1] = 5
slot[2-3] = 6L

这显示了差异代码块中声明的局部变量如何重用插槽.

This shows how the slots are reused by local variables declared in difference code blocks.

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